Search: a268545 -id:a268545
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A005259
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Apery (Apéry) numbers: Sum_{k=0..n} (binomial(n,k)*binomial(n+k,k))^2.
(Formerly M4020)
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+10
127
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1, 5, 73, 1445, 33001, 819005, 21460825, 584307365, 16367912425, 468690849005, 13657436403073, 403676083788125, 12073365010564729, 364713572395983725, 11111571997143198073, 341034504521827105445, 10534522198396293262825, 327259338516161442321485
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OFFSET
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0,2
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COMMENTS
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Conjecture: For each n = 1,2,3,... the Apéry polynomial A_n(x) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
The expansions of exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + 232771*x^5 + ... and exp( Sum_{n >= 1} a(n-1)*x^n/n ) = 1 + 3*x + 27*x^2 + 390*x^3 + 7038*x^4 + 144550*x^5 + ... both appear to have integer coefficients. See A267220. - Peter Bala, Jan 12 2016
Diagonal of the rational function R(x, y, z, w) = 1 / (1 - (w*x*y*z + w*x*y + w*z + x*y + x*z + y + z)); also diagonal of rational function H(x, y, z, w) = 1/(1 - w*(1+x)*(1+y)*(1+z)*(x*y*z + y*z + y + z + 1)). - Gheorghe Coserea, Jun 26 2018
Named after the French mathematician Roger Apéry (1916-1994). - Amiram Eldar, Jun 10 2021
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REFERENCES
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Julian Havil, The Irrationals, Princeton University Press, Princeton and Oxford, 2012, pp. 137-153.
Wolfram Koepf, Hypergeometric Identities. Ch. 2 in Hypergeometric Summation: An Algorithmic Approach to Summation and Special Function Identities. Braunschweig, Germany: Vieweg, pp. 55, 119 and 146, 1998.
Maxim Kontsevich and Don Zagier, Periods, pp. 771-808 of B. Engquist and W. Schmid, editors, Mathematics Unlimited - 2001 and Beyond, 2 vols., Springer-Verlag, 2001.
Leonard Lipshitz and Alfred van der Poorten, "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990), pp. 339-358.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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Roger Apéry, Irrationalité de zeta(2) et zeta(3), in Journées Arith. de Luminy. Colloque International du Centre National de la Recherche Scientifique (CNRS) held at the Centre Universitaire de Luminy, Luminy, Jun 20-24, 1978. Astérisque, Vol. 61 (1979), pp. 11-13.
Frits Beukers, Consequences of Apéry's work on zeta(3), in "Zeta(3) irrationnel: les retombées", Rencontres Arithmétiques de Caen, June 2-3, 1995 [Mentions divisibility of a(n) by powers of 5 and powers of 11]
Leonard Lipshitz and Alfred J. van der Poorten, Rational functions, diagonals, automata and arithmetic, in: Richard A. Mollin (ed.), Number theory, Proceedings of the First Conference of the Canadian Number Theory Association Held at the Banff Center, Banff, Alberta, April 17-27, 1988, de Gruyter, 2016, pp. 339-358; alternative link; Wayback Machine copy.
Alfred van der Poorten, A proof that Euler missed ..., Math. Intelligencer, Vol. 1, No. 4 (December 1979), pp. 196-203, (b_n) after eq. (1.2), and Exercise 3.
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FORMULA
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D-finite with recurrence (n+1)^3*a(n+1) = (34*n^3 + 51*n^2 + 27*n + 5)*a(n) - n^3*a(n-1), n >= 1.
Representation as a special value of the hypergeometric function 4F3, in Maple notation: a(n)=hypergeom([n+1, n+1, -n, -n], [1, 1, 1], 1), n=0, 1, ... - Karol A. Penson Jul 24 2002
G.f.: (-1/2)*(3*x - 3 + (x^2-34*x+1)^(1/2))*(x+1)^(-2)*hypergeom([1/3,2/3],[1],(-1/2)*(x^2 - 7*x + 1)*(x+1)^(-3)*(x^2 - 34*x + 1)^(1/2)+(1/2)*(x^3 + 30*x^2 - 24*x + 1)*(x+1)^(-3))^2. - Mark van Hoeij, Oct 29 2011
Let g(x, y) = 4*cos(2*x) + 8*sin(y)*cos(x) + 5 and let P(n,z) denote the Legendre polynomial of degree n. Then G. A. Edgar posted a conjecture of Alexandru Lupas that a(n) equals the double integral 1/(4*Pi^2)*int {y = -Pi..Pi} int {x = -Pi..Pi} P(n,g(x,y)) dx dy. (Added Jan 07 2015: Answered affirmatively in Math Overflow question 178790) - Peter Bala, Mar 04 2012; edited by G. A. Edgar, Dec 10 2016
a(n) ~ (1+sqrt(2))^(4*n+2)/(2^(9/4)*Pi^(3/2)*n^(3/2)). - Vaclav Kotesovec, Nov 01 2012
a(n) = Sum_{k=0..n} C(n,k)^2 * C(n+k,k)^2. - Joerg Arndt, May 11 2013
0 = (-x^2+34*x^3-x^4)*y''' + (-3*x+153*x^2-6*x^3)*y'' + (-1+112*x-7*x^2)*y' + (5-x)*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j) * C(n,k)^2 * C(n+k,k)^2 * C(n,j) * C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} C(n,k) * C(n+k,k) * C(k,j)^3 (see Koepf, p. 55).
a(n) = Sum_{0 <= j, k <= n} C(n,k)^2 * C(n,j)^2 * C(3*n-j-k,2*n) (see Koepf, p. 119).
Diagonal coefficients of the rational function 1/((1 - x - y)*(1 - z - t) - x*y*z*t) (Straub, 2014). (End)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^m at m = 2. At m = 1 we get the Apéry numbers A005258. - Peter Bala, Dec 22 2020
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k). - Peter Bala, Jul 18 2024
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EXAMPLE
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G.f. = 1 + 5*x + 73*x^2 + 1445*x^3 + 33001*x^4 + 819005*x^5 + 21460825*x^6 + ...
a(2) = (binomial(2,0) * binomial(2+0,0))^2 + (binomial(2,1) * binomial(2+1,1))^2 + (binomial(2,2) * binomial(2+2,2))^2 = (1*1)^2 + (2*3)^2 + (1*6)^2 = 1 + 36 + 36 = 73. - Michael B. Porter, Jul 14 2016
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MAPLE
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a := proc(n) option remember; if n=0 then 1 elif n=1 then 5 else (n^(-3))* ( (34*(n-1)^3 + 51*(n-1)^2 + 27*(n-1) +5)*a((n-1)) - (n-1)^3*a((n-1)-1)); fi; end;
# Alternative:
a := n -> hypergeom([-n, -n, 1+n, 1+n], [1, 1, 1], 1):
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MATHEMATICA
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Table[HypergeometricPFQ[{-n, -n, n+1, n+1}, {1, 1, 1}, 1], {n, 0, 13}] (* Jean-François Alcover, Apr 01 2011 *)
Table[Sum[(Binomial[n, k]Binomial[n+k, k])^2, {k, 0, n}], {n, 0, 30}] (* Harvey P. Dale, Oct 15 2011 *)
a[ n_] := SeriesCoefficient[ SeriesCoefficient[ SeriesCoefficient[ SeriesCoefficient[ 1 / (1 - t (1 + x ) (1 + y ) (1 + z ) (x y z + (y + 1) (z + 1))), {t, 0, n}], {x, 0, n}], {y, 0, n}], {z, 0, n}]; (* Michael Somos, May 14 2016 *)
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PROG
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(Haskell)
a005259 n = a005259_list !! n
a005259_list = 1 : 5 : zipWith div (zipWith (-)
(tail $ zipWith (*) a006221_list a005259_list)
(zipWith (*) (tail a000578_list) a005259_list)) (drop 2 a000578_list)
(GAP) List([0..20], n->Sum([0..n], k->Binomial(n, k)^2*Binomial(n+k, k)^2)); # Muniru A Asiru, Sep 28 2018
(Magma) [&+[Binomial(n, k) ^2 *Binomial(n+k, k)^2: k in [0..n]]:n in [0..17]]; // Marius A. Burtea, Jan 20 2020
(Python)
m, g = 1, 0
for k in range(n+1):
g += m
m *= ((n+k+1)*(n-k))**2
m //=(k+1)**4
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CROSSREFS
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The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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A006480
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De Bruijn's S(3,n): (3n)!/(n!)^3.
(Formerly M4284)
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+10
97
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1, 6, 90, 1680, 34650, 756756, 17153136, 399072960, 9465511770, 227873431500, 5550996791340, 136526995463040, 3384731762521200, 84478098072866400, 2120572665910728000, 53494979785374631680, 1355345464406015082330, 34469858696831179429500, 879619727485803060256500, 22514366432046593564460000
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OFFSET
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0,2
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COMMENTS
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Number of paths of length 3n in an n X n X n grid from (0,0,0) to (n,n,n), using steps (0,0,1), (0,1,0), and (1,0,0).
Appears in Ramanujan's theory of elliptic functions of signature 3.
S(s,n) = Sum_{k=0..2n} (-1)^(k+n) * binomial(2n, k)^s. The formula S(3,n) = (3n)!/(n!)^3 is due to Dixon (according to W. N. Bailey 1935). - Charles R Greathouse IV, Dec 28 2011
a(n) is the number of ballot results that end in a 3-way tie when 3n voters each cast two votes for two out of three candidates vying for 2 slots on a county board; in such a tie, each of the three candidates receives 2n votes. Note there are C(3n,2n) ways to choose the voters who cast a vote for the youngest candidate. The n voters who did note vote for the youngest candidate voted for the two older candidates. Then there are C(2n,n) ways to choose the other n voters who voted for both the youngest and the second youngest candidate. The remaining voters vote for the oldest candidate. Hence there are C(3n,2n)*C(2n,n)=(3n)!/(n!)^3 ballot results. - Dennis P. Walsh, May 02 2013
a(n) is the constant term of (X+Y+1/(X*Y))^(3*n). - Mark van Hoeij, May 07 2013
a(n) is the number of permutations of the multiset {1^n, 2^n, 3^n}, the number of ternary words of length 3*n with n of each letters. - Joerg Arndt, Feb 28 2016
Diagonal of the rational function 1/(1 - x - y - z). - Gheorghe Coserea, Jul 06 2016
At least two families of elliptic curves, x = 2*H1 = (p^2+q^2)*(1-q) and x = 2*H2 = p^2+q^2-3*p^2*q+q^3 (0<x<4/27), generate this sequence via the period-energy function T(x) = 2*Pi*2F1(1/3,2/3; 1; (27/4)*x). - Bradley Klee, Feb 25 2018
The ordinary generating function also determines periods along a family of tetrahedral-symmetric sphere curves ("du troisième ordre"). Compare links to Goursat "Étude des surfaces..." and "Proof Certificate". - Bradley Klee, Sep 28 2018
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REFERENCES
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L. A. Aizenberg and A. P. Yuzhakov, "Integral representations and residues in multidimensional complex analysis", American Mathematical Society, 1983, p. 194.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 174.
N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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Marko Petkovsek, Herbert Wilf and Doron Zeilberger, A=B, A K Peters, 1996, p. 22.
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FORMULA
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Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 1/2 * sqrt(3) * 27^n / (Pi*n) - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
O.g.f.: hypergeom([1/3, 2/3], [1], 27*x).
E.g.f.: hypergeom([1/3, 2/3], [1, 1], 27*x).
Integral representation as n-th moment of a positive function on [0, 27]:
a(n) = int( x^n*(-1/24*(3*sqrt(3)*hypergeom([2/3, 2/3], [4/3], 1/27*x)* Gamma(2/3)^6*x^(1/3) - 8*hypergeom([1/3, 1/3], [2/3], 1/27*x)*Pi^3)/Pi^3 /x^(2/3)/Gamma(2/3)^3), x=0..27). This representation is unique. (End)
a(n) = Sum_{k=-n..n} (-1)^k*binomial(2*n, n+k)^3. - Benoit Cloitre, Mar 02 2005
G.f. satisfies: A(x^3) = A( x*(1+3*x+9*x^2)/(1+6*x)^3 )/(1+6*x). - Paul D. Hanna, Oct 29 2010
D-finite with recurrence: n^2*a(n) - 3*(3*n-1)*(3*n-2)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
0 = a(n)^2*(472392*a(n+1)^2 - 83106*a(n+1)*a(n+2) + 3600*a(n+2)^2) + a(n)*a(n+1)*(-8748*a(n+1)^2 + 1953*a(n+1)*a(n+2) - 120*a(n+2)^2) + a(n+1)^2*(+36*a(n+1)^2 - 12*a(n+1)*a(n+2) + a(n+2)^2 for all n in Z. - Michael Somos, Oct 22 2014
0 = x*(27*x-1)*y'' + (54*x-1)*y' + 6*y, where y is g.f. - Gheorghe Coserea, Jul 06 2016
a(n) = 3*binomial(2*n - 1,n)*binomial(3*n - 1,n) = 3*[x^n] 1/(1 - x)^n * [x^n] 1/(1 - x)^(2*n) for n >= 1.
a(n) = binomial(2*n,n)*binomial(3*n,n) = ([x^n](1 + x)^(2*n)) *([x^n](1 + x)^(3*n)) = [x^n](F(x)^(6*n)), where F(x) = 1 + x + 2*x^2 + 14*x^3 + 127*x^4 + 1364*x^5 + 16219*x^6 + ... appears to have integer coefficients. Cf. A002894.
This sequence occurs as the right-hand side of several binomial sums:
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)^3 = a(n) (Dixon's identity).
Sum_{k = 0..n} binomial(n,k)*binomial(2*n,n - k)*binomial(3*n + k,k) = a(n) (Gould, Vol. 4, 6.86)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n,k)*binomial(2*n + k,n)*binomial(3*n + k,n) = a(n).
Sum_{k = 0..n} binomial(n,k)*binomial(2*n + k,k)*binomial(3*n,n - k) = a(n).
Sum_{k = 0..n} (-1)^(k)*binomial(n,k)*binomial(3*n - k,n)*binomial(4*n - k,n) = a(n).
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n + k,2*n - k)*binomial(2*k,k)*binomial(4*n - k,2*n) = a(n) (see Gould, Vol.5, 9.23).
Sum_{k = 0..2*n} (-1)^k*binomial(3*n,k)*binomial(3*n - k,n)^3 = a(n) (Sprugnoli, Section 2.9, Table 10, p. 123). (End)
G.f.: F(x) = 1/(2*Pi) Integral_{z=0..2*Pi} 2F1(1/3,2/3; 1/2; 27*x*sin^2(z)) dz.
With G(x) = x*2F1(1/3,2/3; 2; 27*x): F(x) = d/dx G(x). (Cf. A007004) (End)
F(x)*G(1/27-x) + F(1/27-x)*G(x) = 1/(4*Pi*sqrt(3)). - Bradley Klee, Sep 29 2018
a(n) = Sum_{k = n..2*n} binomial(2*n,k)^2 * binomial(k,n). Cf. A001459.
a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for any prime p >= 5 and any positive integers n and k (write a(n) as C(3*n,2*n)*C(2*n,n) and apply Mestrovic, equation 39, p. 12). (End)
Occurs on the right-hand side of the binomial sum identities Sum_{k = -n..n} (-1)^k * (n + x - k) * binomial(2*n, n+k)^3 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^3 * binomial(2*n, n+k)^3 = x*(x + n)*(x + 2*n)*a(n) (x arbitrary). Compare with Dixon's identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^3 = a(n). - Peter Bala, Jul 31 2023
a(n) = (-1)^n * [x^(2*n)] ( (1 - x)^(4*n) * Legendre_P(2*n, (1 + x)/(1 - x)) ).
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EXAMPLE
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G.f.: 1 + 6*x + 90*x^2 + 1680*x^3 + 34650*x^4 + 756756*x^5 + 17153136*x^6 + ...
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MAPLE
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MATHEMATICA
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Sum [ (-1)^(k+n) Binomial[ 2n, k ]^3, {k, 0, 2n} ]
a[ n_] := If[ n < 0, 0, (-1)^n HypergeometricPFQ[ {-2 n, -2 n, -2 n}, {1, 1}, 1]]; (* Michael Somos, Oct 22 2014 *)
CoefficientList[Series[Hypergeometric2F1[1/3, 2/3, 1, 27*x], {x, 0, 5}], x] (* Bradley Klee, Feb 28 2018 *)
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PROG
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(PARI) {a(n) = if( n<0, 0, (3*n)! / n!^3)}; /* Michael Somos, Dec 03 2002 */
(PARI) {a(n) = my(A, m); if( n<1, n==0, m=1; A = 1 + O(x); while( m<=n, m*=3; A = subst( (1 + 2*x) * subst(A, x, (x/3)^3), x, serreverse(x * (1 + x + x^2) / (1 + 2*x)^3 / 3 + O(x^m)))); polcoeff(A, n))}; /* Michael Somos, Dec 03 2002 */
(Magma) [Factorial(3*n)/(Factorial(n))^3: n in [0..20] ]; // Vincenzo Librandi, Aug 20 2011
(Maxima) makelist(multinomial_coeff(n, n, n), n, 0, 24); /* Emanuele Munarini, Oct 25 2016 */
(GAP) List([0..20], n->Factorial(3*n)/Factorial(n)^3); # Muniru A Asiru, Mar 31 2018
(Python)
from math import factorial
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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Terms a(17) and beyond from T. D. Noe, Jun 29 2008
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STATUS
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approved
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A268555
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Diagonal of the rational function of six variables 1/((1 - w - u v - u v w) * (1 - z - x y)).
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+10
74
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1, 6, 78, 1260, 22470, 424116, 8305836, 166929048, 3419932230, 71109813060, 1496053026468, 31777397077608, 680354749147164, 14664155597771400, 317877850826299800, 6924815555276838960, 151505459922479997510, 3327336781596164286180
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,2
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COMMENTS
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Also diagonal of rational function R(x,y,z) = 1 /(1 - x - y - z - x*y + x*z).
Annihilating differential operator: x*(16*x^2-24*x+1)*(d/dx)^2 + (48*x^2-48*x+1)*(d/dx) + 12*x-6.
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LINKS
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FORMULA
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D-finite with recurrence: n^2*a(n) -6*(2*n-1)^2*a(n-1) +4*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Mar 10 2016
a(n) ~ sqrt(4+3*sqrt(2)) * 2^(2*n-3/2) * (1+sqrt(2))^(2*n) / (Pi*n). - Vaclav Kotesovec, Jul 01 2016
G.f.: hypergeom([1/12, 5/12],[1],6912*x^4*(1-24*x+16*x^2)/(1-24*x+48*x^2)^3)/(1-24*x+48*x^2)^(1/4).
0 = x*(16*x^2-24*x+1)*y'' + (48*x^2-48*x+1)*y' + (12*x-6)*y, where y is g.f.
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EXAMPLE
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G.f. = 1 + 6*x + 78*x^2 + 1260*x^3 + 22470*x^4 + 424116*x^5 + 8305836*x^6 + ...
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MAPLE
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1/(1-w-u*v-u*v*w)/(1-z-x*y) ;
coeftayl(%, x=0, n) ;
coeftayl(%, y=0, n) ;
coeftayl(%, z=0, n) ;
coeftayl(%, u=0, n) ;
coeftayl(%, v=0, n) ;
coeftayl(%, w=0, n) ;
end proc:
seq(binomial(2*n, n)*add(binomial(n, k)*binomial(n+k, k), k = 0..n), n = 0..20); # Peter Bala, Mar 19 2018
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MATHEMATICA
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sc = SeriesCoefficient;
a[n_] := 1/(1-w-u*v-u*v*w)/(1-z-x*y) // sc[#, {x, 0, n}]& // sc[#, {y, 0, n}]& // sc[#, {z, 0, n}]& // sc[#, {u, 0, n}]& // sc[#, {v, 0, n}]& // sc[#, {w, 0, n}]&;
a[n_] := Product[Hypergeometric2F1[-n, -n, 1, i], {i, 1, 2}];
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PROG
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(PARI) \\ system("wget http://www.jjj.de/pari/hypergeom.gpi");
read("hypergeom.gpi");
N = 18; x = 'x + O('x^N);
Vec(hypergeom_sym([1/12, 5/12], [1], 6912*x^4*(1-24*x+16*x^2)/(1-24*x+48*x^2)^3, N)/(1-24*x+48*x^2)^(1/4)) \\ Gheorghe Coserea, Jul 05 2016
(PARI) {a(n) = if( n<1, n==0, my(A = vector(n+1)); A[1] = 1; A[2] = 6; for(k=2, n, A[k+1] = (6*(2*k-1)^2*A[k] - 4*(2*k-1)*(2*k-3)*A[k-1]) / k^2); A[n+1])}; /* Michael Somos, Jan 22 2017 */
(PARI)
diag(expr, N=22, var=variables(expr)) = {
my(a = vector(N));
for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
for (n = 1, N, a[n] = expr;
for (k = 1, #var, a[n] = polcoeff(a[n], n-1)));
return(a);
};
diag(1/(1 - x - y - z - x*y + x*z), 18)
\\ test: diag(1/(1-x-y-z-x*y+x*z)) == diag(1/((1-w-u*v-u*v*w)*(1-z-x*y)))
(GAP) List([0..20], n->Binomial(2*n, n)*Sum([0..n], k->Binomial(n, k)*Binomial(n+k, k))); # Muniru A Asiru, Mar 19 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A005260
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a(n) = Sum_{k = 0..n} binomial(n,k)^4.
(Formerly M2110)
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+10
65
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1, 2, 18, 164, 1810, 21252, 263844, 3395016, 44916498, 607041380, 8345319268, 116335834056, 1640651321764, 23365271704712, 335556407724360, 4854133484555664, 70666388112940818, 1034529673001901732
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OFFSET
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0,2
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COMMENTS
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Diagonal of the rational function R(x,y,z,w) = 1/(1 - (w*x*y + w*x*z + w*y*z + x*y*z + w*x + y*z)). - Gheorghe Coserea, Jul 13 2016
This is one of the Apéry-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Every prime eventually divides some term of this sequence. - Amita Malik, Aug 20 2017
Two walkers, A and B, stand on the South-West and North-East corners of an n X n grid, respectively. A walks by either North or East steps while B walks by either South or West steps. Sequence values a(n) < binomial(2*n,n)^2 count the simultaneous walks where A and B meet after exactly n steps and change places after 2*n steps. - Bradley Klee, Apr 01 2019
a(n) is the constant term in the expansion of ((1 + x) * (1 + y) * (1 + z) + (1 + 1/x) * (1 + 1/y) * (1 + 1/z))^n. - Seiichi Manyama, Oct 27 2019
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REFERENCES
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H. W. Gould, Combinatorial Identities, Morgantown, 1972, (X.14), p. 79.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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a(n) ~ 2^(1/2)*Pi^(-3/2)*n^(-3/2)*2^(4*n). - Joe Keane (jgk(AT)jgk.org), Jun 21 2002
D-finite with recurrence: n^3*a(n) = 2*(2*n - 1)*(3*n^2 - 3*n + 1)*a(n-1) + (4*n - 3)*(4*n - 4)*(4*n - 5)*a(n-2). [Yuan]
G.f.: 5*hypergeom([1/8, 3/8],[1], (4/5)*((1-16*x)^(1/2)+(1+4*x)^(1/2))*(-(1-16*x)^(1/2)+(1+4*x)^(1/2))^5/(2*(1-16*x)^(1/2)+3*(1+4*x)^(1/2))^4)^2/(2*(1-16*x)^(1/2)+3*(1+4*x)^(1/2)). - Mark van Hoeij, Oct 29 2011
1/Pi = sqrt(15)/18 * Sum_{n >= 0} a(n)*(4*n + 1)/36^n (Cooper, equation (5)) = sqrt(15)/18 * Sum_{n >= 0} a(n)*A016813(n)/A009980(n). - Jason Kimberley, Nov 26 2012
0 = (-x^2 + 12*x^3 + 64*x^4)*y''' + (-3*x + 54*x^2 + 384*x^3)*y'' + (-1 + 40*x + 444*x^2)*y' + (2 + 60*x)*y, where y is g.f. - Gheorghe Coserea, Jul 13 2016
For r a nonnegative integer, Sum_{k = r..n} C(k,r)^4*C(n,k)^4 = C(n,r)^4*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016
a(n) = hypergeom([-n, -n, -n, -n], [1, 1, 1], 1). - Peter Luschny, Jul 27 2016
Sum_{n>=0} a(n) * x^n / (n!)^4 = (Sum_{n>=0} x^n / (n!)^4)^2. - Ilya Gutkovskiy, Jul 17 2020
a(n) = Sum_{k=0..n} C(n,k)*C(n+k,k)*C(2k,k)*C(2n-2k,n-k)*(-1)^(n-k). This can be proved via the Zeilberger algorithm. - Zhi-Wei Sun, Aug 23 2020
a(n) = (-1)^n*binomial(2*n, n)*hypergeom([1/2, -n, -n, n + 1], [1, 1, 1/2 - n], 1). - Peter Luschny, Aug 24 2020
a(n) = Sum_{k=0..n} binomial(n,k)^2*binomial(2*k,n)*binomial(2*n-k,n) [Theorem 1 in Belbachir and Otmani]. - Michel Marcus, Dec 06 2020
a(n) = [x^n] (1 - x)^(2*n) P(n,(1 + x)/(1 - x))^2, where P(n,x) denotes the n-th Legendre polynomial. See Gould, p. 66. This formula is equivalent to the binomial sum identity of Zhi-Wei Sun given above. - Peter Bala, Mar 24 2022
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EXAMPLE
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G.f. = 1 + 2*x + 18*x^2 + 164*x^3 + 1810*x^4 + 21252*x^5 + 263844*x^6 + ...
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MAPLE
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add( (binomial(n, k))^4, k=0..n) ;
end proc:
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MATHEMATICA
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Table[Sum[Binomial[n, k]^4, {k, 0, n}], {n, 0, 20}] (* Wesley Ivan Hurt, Mar 09 2014 *)
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PROG
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(PARI) {a(n) = sum(k=0, n, binomial(n, k)^4)};
(Python)
m, g = 1, 0
for k in range(n+1):
g += m
m = m*(n-k)**4//(k+1)**4
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CROSSREFS
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The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A006077
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(n+1)^2*a(n+1) = (9n^2+9n+3)*a(n) - 27*n^2*a(n-1), with a(0) = 1 and a(1) = 3.
(Formerly M2775)
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+10
51
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1, 3, 9, 21, 9, -297, -2421, -12933, -52407, -145293, -35091, 2954097, 25228971, 142080669, 602217261, 1724917221, 283305033, -38852066421, -337425235479, -1938308236731, -8364863310291, -24286959061533, -3011589296289, 574023003011199, 5028616107443691
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OFFSET
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0,2
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COMMENTS
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This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
Conjecture: Let W(n) be the (n+1) X (n+1) Hankel-type determinant with (i,j)-entry equal to a(i+j) for all i,j = 0,...,n. If n == 1 (mod 3) then W(n) = 0. When n == 0 or 2 (mod 3), W(n)*(-1)^(floor((n+1)/3))/6^n is always a positive odd integer. - Zhi-Wei Sun, Aug 21 2013
Conjecture: Let p == 1 (mod 3) be a prime, and write 4*p = x^2 + 27*y^2 with x, y integers and x == 1 (mod 3). Then W(p-1) == (-1)^{(p+1)/2}*(x-p/x) (mod p^2), where W(n) is defined as the above. - Zhi-Wei Sun, Aug 23 2013
This is one of the Apery-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Diagonal of rational functions 1/(1 - (x^2*y + y^2*z - z^2*x + 3*x*y*z)), 1/(1 - (x^3 + y^3 - z^3 + 3*x*y*z)), 1/(1 + x^3 + y^3 + z^3 - 3*x*y*z). - Gheorghe Coserea, Aug 04 2018
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REFERENCES
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Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
D. Zagier, Integral solutions of Apery-like recurrence equations, in: Groups and Symmetries: from Neolithic Scots to John McKay, CRM Proc. Lecture Notes 47, Amer. Math. Soc., Providence, RI, 2009, pp. 349-366.
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LINKS
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FORMULA
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G.f.: hypergeom([1/3, 2/3], [1], x^3/(x-1/3)^3) / (1-3*x). - Mark van Hoeij, Oct 25 2011
a(n) = Sum_{k=0..floor(n/3)}(-1)^k*3^(n-3k)*C(n,3k)*C(2k,k)*C(3k,k). - Zhi-Wei Sun, Aug 21 2013
0 = x*(x^2+9*x+27)*y'' + (3*x^2 + 18*x + 27)*y' + (x + 3)*y, where y(x) = A(x/-27). - Gheorghe Coserea, Aug 26 2016
a(n) = 3^n*hypergeom([-n/3, (1-n)/3, (2-n)/3], [1, 1], 1). - Peter Luschny, Nov 01 2017
The g.f. T(x) obeys a period-annihilating ODE:
0=3*(-1 + 9*x)*T(x) + (-1 + 9*x)^2*T'(x) + x*(1 - 9*x + 27*x^2)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = 3*(-8 + 9*(1 - 9*x)^3)*(1 - 9*x);
g3 = 8 - 36*(1 - 9*x)^3 + 27*(1 - 9*x)^6;
which determine an elliptic surface with four singular fibers. (End)
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EXAMPLE
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G.f. = 1 + 3*x + 9*x^2 + 21*x^3 + 9*x^4 - 297*x^5 - 2421*x^6 - 12933*x^7 - ...
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MAPLE
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a := n -> 3^n*hypergeom([-n/3, (1-n)/3, (2-n)/3], [1, 1], 1):
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MATHEMATICA
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Table[Sum[(-1)^k*3^(n - 3*k)*Binomial[n, 3*k]*Binomial[2*k, k]* Binomial[3*k, k], {k, 0, Floor[n/3]}], {n, 0, 50}] (* G. C. Greubel, Oct 24 2017 *)
a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/3, 2/3}, {1}, x^3 / (x - 1/3)^3 ] / (1 - 3 x), {x, 0, n}]; (* Michael Somos, Nov 01 2017 *)
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PROG
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(PARI) subst(eta(q)^3/eta(q^3), q, serreverse(eta(q^9)^3/eta(q)^3*q)) \\ (generating function) Helena Verrill (verrill(AT)math.lsu.edu), Apr 20 2009 [for (-1)^n*a(n)]
(PARI)
diag(expr, N=22, var=variables(expr)) = {
my(a = vector(N));
for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
for (n = 1, N, a[n] = expr;
for (k = 1, #var, a[n] = polcoeff(a[n], n-1)));
return(a);
};
diag(1/(1 + x^3 + y^3 + z^3 - 3*x*y*z), 25)
(PARI)
seq(N) = {
my(a = vector(N)); a[1] = 3; a[2] = 9;
for (n = 2, N-1, a[n+1] = ((9*n^2+9*n+3)*a[n] - 27*n^2*a[n-1])/(n+1)^2);
concat(1, a);
};
seq(24)
\\ test: y=subst(Ser(seq(202)), 'x, -'x/27); 0 == x*(x^2+9*x+27)*y'' + (3*x^2+18*x+27)*y' + (x+3)*y
(PARI) {a(n) = my(A); if( n<0, 0, A = x * O(x^n); (-1)^n * polcoeff( subst(eta(x + A)^3 / eta(x^3 + A), x, serreverse( x * eta(x^9 + A)^3 / eta(x + A)^3)), n))}; /* Michael Somos, Nov 01 2017 */
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CROSSREFS
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The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.
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KEYWORD
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sign
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AUTHOR
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EXTENSIONS
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More terms from Kok Seng Chua (chuaks(AT)ihpc.nus.edu.sg), Jun 20 2000
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STATUS
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approved
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A008977
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a(n) = (4*n)!/(n!)^4.
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+10
46
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1, 24, 2520, 369600, 63063000, 11732745024, 2308743493056, 472518347558400, 99561092450391000, 21452752266265320000, 4705360871073570227520, 1047071828879079131681280, 235809301462142612780721600, 53644737765488792839237440000, 12309355935372581458927646400000
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OFFSET
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0,2
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COMMENTS
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Number of paths of length 4*n in an n X n X n X n grid from (0,0,0,0) to (n,n,n,n).
a(n) occurs in Ramanujan's formula 1/Pi = (sqrt(8)/9801) * Sum_{n>=0} (4*n)!/(n!)^4 * (1103 + 26390*n)/396^(4*n) ). - Susanne Wienand, Jan 05 2013
a(n) is the number of ballot results that lead to a 4-way tie when 4*n voters each cast three votes for three out of four candidates vying for 3 slots on a county commission; each of these ballot results give 3*n votes to each of the four candidates. - Dennis P. Walsh, May 02 2013
a(n) is the constant term of (X + Y + Z + 1/(X*Y*Z))^(4*n). - Mark van Hoeij, May 07 2013
In Narumiya and Shiga on page 158 the g.f. is given as a hypergeometric function. - Michael Somos, Aug 12 2014
Diagonal of the rational function R(x,y,z,w) = 1/(1-(w+x+y+z)). - Gheorghe Coserea, Jul 15 2016
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LINKS
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N. Narumiya and H. Shiga, The mirror map for a family of K3 surfaces induced from the simplest 3-dimensional reflexive polytope, Proceedings on Moonshine and related topics (Montréal, QC, 1999), 139-161, CRM Proc. Lecture Notes, 30, Amer. Math. Soc., Providence, RI, 2001. MR1877764 (2002m:14030).
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FORMULA
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Self-convolution of A178529, where A178529(n) = (4^n/n!^2) * Product_{k=0..n-1} (8*k + 1)*(8*k + 3).
G.f.: hypergeom([1/8, 3/8], [1], 256*x)^2. - Mark van Hoeij, Nov 16 2011
G.f.: hypergeom([1/4, 2/4, 3/4], [1, 1], 256*x). - Michael Somos, Aug 12 2014
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(24*n)), where F(x) = 1 + x + 29*x^2 + 2246*x^3 + 239500*x^4 + 30318701*x^5 + 4271201506*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008978, A008979, A186420 and A188662. (End)
0 = (x^2-256*x^3)*y''' + (3*x-1152*x^2)*y'' + (1-816*x)*y' - 24*y, where y is the g.f. - Gheorghe Coserea, Jul 15 2016
a(n) = Sum_{k = 0..3*n} (-1)^(n+k)*binomial(4*n,n + k)* binomial(n + k,k)^4.
a(n) = Sum_{k = 0..4*n} (-1)^k*binomial(4*n,k)*binomial(n + k,k)^4. (End)
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z)^n] (1 + x + y + z)^(4*n). (End)
D-finite with recurrence n^3*a(n) -8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1)=0. - R. J. Mathar, Aug 01 2022
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EXAMPLE
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a(13)=52!/(13!)^4=53644737765488792839237440000 is the number of ways of dealing the four hands in Bridge or Whist. - Henry Bottomley, Oct 06 2000
a(1)=24 since, in a 4-voter 3-vote election that ends in a four-way tie for candidates A, B, C, and D, there are 4! ways to arrange the needed vote sets {A,B,C}, {A,B,D}, {A,C,D}, and {B,C,D} among the 4 voters. - Dennis P. Walsh, May 02 2013
G.f. = 1 + 24*x + 2520*x^2 + 369600*x^3 + 63063000*x^4 + 11732745024*x^5 + ...
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MAPLE
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MATHEMATICA
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a[ n_] := If[ n < 0, 0, (4 n)! / n!^4]; (* Michael Somos, Aug 12 2014 *)
a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/4, 2/4, 3/4}, {1, 1}, 256 x], {x, 0, n}]; (* Michael Somos, Aug 12 2014 *)
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PROG
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(Magma) [Factorial(4*n)/Factorial(n)^4: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014
(Python)
from math import factorial
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CROSSREFS
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Cf. A000984, A006480, A008978, A178529, A000897, A002894, A002897, A006480, A008979, A186420, A188662.
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KEYWORD
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nonn,easy,nice
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AUTHOR
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STATUS
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approved
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1, 4, 48, 760, 13840, 273504, 5703096, 123519792, 2751843600, 62659854400, 1451780950048, 34116354472512, 811208174862904, 19481055861877120, 471822589361293680, 11511531876280913760, 282665135367572129040
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OFFSET
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0,2
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COMMENTS
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The g.f. for row n of triangle A181544 is (1-x)^(3n+1)*Sum_{k>=0}C(n+k-1,k)^3*x^k.
Diagonal of the rational function R(x,y,z,w) = 1/(1 - (w*x*y + w*x*z + w*y*z + x*y + x*z + y + z)). - Gheorghe Coserea, Jul 14 2016
This is one of the Apery-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Every prime eventually divides some term of this sequence. - Amita Malik, Aug 20 2017
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LINKS
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FORMULA
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a(n) = [x^n] (1-x)^(3n+1) * Sum_{k>=0} C(n+k-1,k)^3*x^k.
a(n) = Sum_{j = 0..n} C(n,j)^2 * C(2*j,n) * C(j+n,j). [Formula of Wadim Zudilin provided by Jason Kimberley, Nov 06 2012]
1/Pi = sqrt(7) Sum_{n>=0} (-1)^n a(n) (11895n + 1286)/22^(3n+3). [Cooper, equation (41)] - Jason Kimberley, Nov 06 2012
G.f.: sqrt((1-13*x+(1-26*x-27*x^2)^(1/2))/(1-21*x+8*x^2+(1-8*x)*(1-26*x-27*x^2)^(1/2)))*hypergeom([1/12,5/12],[1],13824*x^7/(1-21*x+8*x^2+(1-8*x)*(1-26*x-27*x^2)^(1/2))^3)^2. - Mark van Hoeij, May 07 2013
G.f. A(x) satisfies 1/(1+4*x)^2 * A( x/(1+4*x)^3 ) = 1/(1+2*x)^2 * A( x^2/(1+2*x)^3 ) [see Cooper, Guillera, Straub, Zudilin]. - Joerg Arndt, Apr 08 2016
a(n) = (-1)^n*binomial(3n+1,n)* 4F3({-n,n+1,n+1,n+1};{1,1,2(n+1)}; 1). - M. Lawrence Glasser, May 15 2016
Conjecture D-finite with recurrence: n^3*a(n) - (2*n-1)*(13*n^2-13*n+4)*a(n-1) - 3*(n-1)*(3*n-4)*(3*n-2)*a(n-2) = 0. - R. J. Mathar, May 15 2016
0 = (-x^2+26*x^3+27*x^4)*y''' + (-3*x+117*x^2+162*x^3)*y'' + (-1+86*x+186*x^2)*y' + (4+24*x)*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
The conjectured D-finite recurrence can be proved by Zeilberger's algorithm.
a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(n+k,n) * binomial(2*n-k,n) = [(w*x*y*z)^n] ((w+y)*(x+z)*(y+z)*(w+x+y+z))^n. (End)
a(n) = Sum_{0 <= j, k <= n} binomial(n, k)^2 * binomial(n, j)^2 * binomial(k+j, n) = Sum_{k = 0..n} binomial(n, k)^2 * A108625(n, k). - Peter Bala, Jul 08 2024
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EXAMPLE
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(1);
1, (4), 1;
1, 20, (48), 20, 1;
1, 54, 405, (760), 405, 54, 1;
1, 112, 1828, 8464, (13840), 8464, 1828, 112, 1; ...
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MATHEMATICA
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Table[Sum[Binomial[n, j]^2 * Binomial[2*j, n] * Binomial[j+n, j], {j, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 05 2015 *)
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PROG
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(PARI) {a(n)=polcoeff((1-x)^(3*n+1)*sum(j=0, 2*n, binomial(n+j, j)^3*x^j), n)}
(Magma) P<x>:=PolynomialRing(Integers()); C:=Binomial;
A183204:=func<n|Coefficient((1-x)^(3*n+1)*&+[C(n+j, j)^3*x^j:j in[0..2*n]], n)>; // or directly:
A183204:=func<k|&+[C(k, j)^2*C(2*j, k)*C(j+k, j):j in[0..k]]>;
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CROSSREFS
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The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A002897
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a(n) = binomial(2n,n)^3.
(Formerly M4580 N1952)
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+10
34
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1, 8, 216, 8000, 343000, 16003008, 788889024, 40424237568, 2131746903000, 114933031928000, 6306605327953216, 351047164190381568, 19774031697705428416, 1125058699232216000000, 64561313052442296000000
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OFFSET
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0,2
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COMMENTS
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Diagonal of the rational function R(x,y,z,w) = 1/(1 - (w*x*y + w*z + x + y + z)). - Gheorghe Coserea, Jul 14 2016
Conjecture: The g.f. is also the diagonal of the rational function 1/(1 - (x + y)*(1 - 4*z*t) - z - t) = 1/det(I - M*diag(x, y, z, t)), I the 4 x 4 unit matrix and M the 4 x 4 matrix [1, 1, 1, 1; 1, 1, 1, 1; 1, 1, 1, -1; 1 , 1, -1, 1]. If true, then a(n) = [(x*y*z)^n] (1 + x + y + z)^(2*n)*(1 + x + y - z)^n*(1 + x - y + z)^n. - Peter Bala, Apr 10 2022
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REFERENCES
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S. Ramanujan, Modular Equations and Approximations to pi, pp. 23-39 of Collected Papers of Srinivasa Ramanujan, Ed. G. H. Hardy et al., AMS Chelsea 2000. See page 36, equation (25).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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Expansion of (K(k)/(Pi/2))^2 in powers of (kk'/4)^2, where K(k) is the complete elliptic integral of the first kind evaluated at modulus k. - Michael Somos, Jan 31 2007
G.f.: F(1/2, 1/2, 1/2; 1, 1; 64x) where F() is a hypergeometric function. - Michael Somos, Jan 31 2007
D-finite with recurrence n^3*a(n) - 8*(2*n - 1)^3*a(n-1) = 0. - R. J. Mathar, Mar 08 2013
a(n) = binomial(2*n,n)^3 = ( [x^n](1 + x)^(2*n) )^3 = [x^n](F(x)^(8*n)), where F(x) = 1 + x + 6*x^2 + 111*x^3 + 2806*x^4 + 84456*x^5 + 2832589*x^6 + 102290342*x^7 + ... appears to have integer coefficients. For similar results see A000897, A002894, A006480, A008977, A186420 and A188662. (End)
0 = (-x^2 + 64*x^3)*y''' + (-3*x + 288*x^2)*y'' + (-1 + 208*x)*y' + 8*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
a(n) = Sum_{k = 0..n} (2*n + k)!/(k!^3*(n - k)!^2). Cf. A001850(n) = Sum_{k = 0..n} (n + k)!/(k!^2*(n - k)!). - Peter Bala, Jul 27 2016
It appears that a(n) is the coefficient of (x*y*z)^(2*n) in the expansion of (1 + x*y + x*z - y*z)^(2*n) * (1 + x*y - x*z + y*z)^(2*n) * (1 - x*y + x*z + y*z)^(2*n). Cf. A000172. - Peter Bala, Sep 21 2021
a(n) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)*binomial(2*n+k,n).
a(n) = the coefficient of (x*y*z*t^2)^n in the expansion of 1/(1 - x - y)*(1 - z - t) - x*y*z*t) (a(n) = A(n,n,n,2*n) in the notation of Straub, Theorem 1.2). (End)
a(n) = (8/5) * Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)*binomial(2*n+k-1,n) for n >= 1. - Peter Bala, Jul 09 2024
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MATHEMATICA
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a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/2, 1/2, 1/2}, {1, 1}, 64x], {x, 0, n}];
Table[Binomial[2n, n]^3, {n, 0, 20}] (* Harvey P. Dale, Dec 06 2017 *)
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PROG
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(PARI) {a(n) = binomial(2*n, n)^3}; /* Michael Somos, Jan 31 2007 */
(Sage) [binomial(2*n, n)**3 for n in range(21)] # Zerinvary Lajos, Apr 21 2009
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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A004987
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a(n) = (3^n/n!)*Product_{k=0..n-1} (3*k + 1).
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+10
33
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1, 3, 18, 126, 945, 7371, 58968, 480168, 3961386, 33011550, 277297020, 2344420260, 19927572210, 170150808870, 1458435504600, 12542545339560, 108179453553705, 935434098376155, 8107095519260010, 70403724246205350, 612512400941986545, 5337608065351597035, 46582761297613937760
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OFFSET
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0,2
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COMMENTS
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Diagonal of rational function R(x,y) = (1 - 9*x*y) / (1 - 2*x - 3*y + 3*y^2 + 9*x^2*y). - Gheorghe Coserea, Jul 01 2016
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LINKS
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FORMULA
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G.f.: (1 - 9*x)^(-1/3).
a(n) = (3^n/n!)*A007559(n), n >= 1, a(0) := 1.
a(n) ~ Gamma(1/3)^-1*n^(-2/3)*3^(2*n)*{1 - 1/9*n^-1 + ...}.
Representation as n-th moment of a positive function on (0, 9): a(n) = Integral_{x=0..9} ( x^n*(1/(Pi*sqrt(3)*6*(x/9)^(2/3)*(1-x/9)^(1/3))) ), n >= 0. This function is the solution of the Hausdorff moment problem on (0, 9) with moments equal to a(n). As a consequence this representation is unique. - Karol A. Penson, Jan 30 2003
D-finite with recurrence: n*a(n) + 3*(2-3*n)*a(n-1)=0. - R. J. Mathar, Jun 07 2013
0 = a(n) * (81*a(n+1) - 15*a(n+2)) + a(n+1) * (-3*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Jan 27 2014
G.f. A(x)=:y satisfies 0 = y'' * y - 4 * y' * y'. - Michael Somos, Jan 27 2014
E.g.f.: is the hypergeometric function of type 1F1, in Maple notation hypergeom([1/3], [1], 9*x). - Karol A. Penson, Dec 19 2015
Sum_{n>=0} 1/a(n) = (sqrt(3)*Pi + 3*(12 + log(3)))/32 = 1.3980385924595932... - Ilya Gutkovskiy, Jul 01 2016
a(n) = (9^n)*Sum_{k = 0..2*n} (-1)^k*binomial(-1/3, k)* binomial(-1/3, 2*n - k).
(9^n)*a(n) = Sum_{k = 0..2*n} (-1)^k*a(k)*a(2*n-k).
Sum_{k = 0..n} a(k)*a(n-k) = A004988(n).
Sum_{k = 0..2*n} a(k)*a(2*n-k) = 18^n/(2*n)! * Product_{k = 1..n} (6*k - 1)*(3*k - 2). (End)
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EXAMPLE
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G.f.: 1 + 3*x + 18*x^2 + 126*x^3 + 945*x^4 + 7371*x^5 + 58968*x^6 + 480168*x^7 + ...
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MAPLE
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a:= n-> (3^n/n!)*mul(3*k+1, k=0..n-1); seq(a(n), n=0..25); # G. C. Greubel, Aug 22 2019
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MATHEMATICA
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PROG
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(PARI) a(n) = prod(k=0, n-1, 3*k + 1)*3^n/n! \\ Michel Marcus, Jun 30 2013
(PARI)
my(x='x, y='y);
R = (1 - 9*x*y) / (1 - 2*x - 3*y + 3*y^2 + 9*x^2*y);
diag(n, expr, var) = {
my(a = vector(n));
for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n));
for (k = 1, n, a[k] = expr;
for (i = 1, #var, a[k] = polcoeff(a[k], k-1)));
return(a);
};
(Magma) [1] cat [3^n*&*[3*k+1: k in [0..n-1]]/Factorial(n): n in [1..25]]; // G. C. Greubel, Aug 22 2019
(Sage) [9^n*rising_factorial(1/3, n)/factorial(n) for n in (0..25)] # G. C. Greubel, Aug 22 2019
(GAP) List([0..25], n-> 3^n*Product([0..n-1], k-> 3*k+1)/Factorial(n) ); # G. C. Greubel, Aug 22 2019
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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Joe Keane (jgk(AT)jgk.org)
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EXTENSIONS
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STATUS
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approved
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A002896
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Number of 2n-step polygons on cubic lattice.
(Formerly M4285 N1791)
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+10
32
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1, 6, 90, 1860, 44730, 1172556, 32496156, 936369720, 27770358330, 842090474940, 25989269017140, 813689707488840, 25780447171287900, 825043888527957000, 26630804377937061000, 865978374333905289360, 28342398385058078078010, 932905175625150142902300
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OFFSET
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0,2
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COMMENTS
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Number of walks with 2n steps on the cubic lattice Z^3 beginning and ending at (0,0,0).
If A is a random matrix in USp(6) (6 X 6 complex matrices that are unitary and symplectic) then a(n) is the 2n-th moment of tr(A^k) for all k >= 7. - Andrew V. Sutherland, Mar 24 2008
Diagonal of the rational function R(x,y,z,w) = 1/(1 - (w*x*y + w*x*z + w*y + x*z + y + z)). - Gheorghe Coserea, Jul 14 2016
Constant term in the expansion of (x + 1/x + y + 1/y + z + 1/z)^(2n). - Harry Richman, Apr 29 2020
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REFERENCES
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N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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a(n) = C(2*n, n)*Sum_{k=0..n} C(n, k)^2*C(2*k, k).
a(n) = (4^n*p(1/2, n)/n!)*hypergeom([-n, -n, 1/2], [1, 1], 4), where p(a, k) = Product_{i=0..k-1} (a+i).
E.g.f.: Sum_{n>=0} a(n)*x^(2*n)/(2*n)! = BesselI(0, 2*x)^3. - Corrected by Christopher J. Smyth, Oct 29 2012
D-finite with recurrence: n^3*a(n) = 2*(2*n-1)*(10*n^2-10*n+3)*a(n-1) - 36*(n-1)*(2*n-1)*(2*n-3)*a(n-2). - Vladeta Jovovic, Jul 16 2004
An asymptotic formula follows immediately from an observation of Bruce Richmond and me in SIAM Review - 31 (1989, 122-125). We use Hayman's method to find the asymptotic behavior of the sum of squares of the multinomial coefficients multi(n, k_1, k_2, ..., k_m) with m fixed. From this one gets a_n ~ (3/4)*sqrt(3)*6^(2*n)/(Pi*n)^(3/2). - Cecil C Rousseau (ccrousse(AT)memphis.edu), Mar 14 2006
G.f.: (1/sqrt(1+12*z)) * hypergeom([1/8,3/8],[1],64/81*z*(1+sqrt(1-36*z))^2*(2+sqrt(1-36*z))^4/(1+12*z)^4) * hypergeom([1/8, 3/8],[1],64/81*z*(1-sqrt(1-36*z))^2*(2-sqrt(1-36*z))^4/(1+12*z)^4). - Sergey Perepechko, Jan 26 2011
G.f.: (1/2)*(10-72*x-6*(144*x^2-40*x+1)^(1/2))^(1/2)*hypergeom([1/6, 1/3],[1],54*x*(108*x^2-27*x+1+(9*x-1)*(144*x^2-40*x+1)^(1/2)))^2. - Mark van Hoeij, Nov 12 2011
0 = (-x^2+40*x^3-144*x^4)*y''' + (-3*x+180*x^2-864*x^3)*y'' + (-1+132*x-972*x^2)*y' + (6-108*x)*y, where y is the g.f. - Gheorghe Coserea, Jul 14 2016
a(n) = (1/Pi)^3*Integral_{0 <= x, y, z <= Pi} (2*cos(x) + 2*cos(y) + 2*cos(z))^(2*n) dx dy dz. - Peter Bala, Feb 10 2022
a(n) = Sum_{i+j+k=n, 0<=i,j,k<=n} multinomial(2n [i,i,j,j,k,k]). - Shel Kaphan, Jan 16 2023
Sum_{k>=0} a(k)/36^k = A086231 = (sqrt(3)-1) * (Gamma(1/24) * Gamma(11/24))^2 / (32*Pi^3). - Vaclav Kotesovec, Apr 23 2023
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EXAMPLE
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1 + 6*x + 90*x^2 + 1860*x^3 + 44730*x^4 + 1172556*x^5 + 32496156*x^6 + ...
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MAPLE
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a := proc(n) local k; binomial(2*n, n)*add(binomial(n, k)^2 *binomial(2*k, k), k=0..n); end;
# second Maple program
a:= proc(n) option remember; `if`(n<2, 5*n+1,
(2*(2*n-1)*(10*n^2-10*n+3) *a(n-1)
-36*(n-1)*(2*n-1)*(2*n-3) *a(n-2)) /n^3)
end:
A002896 := n -> binomial(2*n, n)*hypergeom([1/2, -n, -n], [1, 1], 4):
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MATHEMATICA
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f[n_] := 4^n*Gamma[n + 1/2]*Sum[Binomial[n, k]^2 Binomial[2 k, k], {k, 0, n}]/(Sqrt[Pi]*n!); Array[f, 17, 0] (* Robert G. Wilson v, Oct 29 2011 *)
Table[Binomial[2n, n]Sum[Binomial[n, k]^2 Binomial[2k, k], {k, 0, n}], {n, 0, 20}] (* Harvey P. Dale, Jan 24 2012 *)
a[ n_] := If[ n < 0, 0, HypergeometricPFQ[ {-n, -n, 1/2}, {1, 1}, 4] Binomial[ 2 n, n]] (* Michael Somos, May 21 2013 *)
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PROG
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(PARI) a(n)=binomial(2*n, n)*sum(k=0, n, binomial(n, k)^2*binomial(2*k, k)) \\ Charles R Greathouse IV, Oct 31 2011
(Sage)
x, y, n = 1, 6, 1
while True:
yield x
n += 1
x, y = y, ((4*n-2)*((10*(n-1)*n+3)*y-18*(n-1)*(2*n-3)*x))//n^3
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CROSSREFS
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KEYWORD
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nonn,easy,walk,nice,changed
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AUTHOR
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STATUS
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approved
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