|
|
A052529
|
|
Expansion of (1-x)^3/(1 - 4*x + 3*x^2 - x^3).
|
|
22
|
|
|
1, 1, 4, 13, 41, 129, 406, 1278, 4023, 12664, 39865, 125491, 395033, 1243524, 3914488, 12322413, 38789712, 122106097, 384377665, 1209982081, 3808901426, 11990037126, 37743426307, 118812495276, 374009739309, 1177344897715
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
a(n+1) is the number of distinct matrix products in (A+B+C+D)^n where commutator [A,B] = [A,C] = [B,C] = 0 but D does not commute with A, B, or C. - Paul D. Hanna and Max Alekseyev, Feb 01 2006
Starting (1, 4, 13, ...) = INVERT transform of the triangular series, (1, 3, 6, 10, ...). Example: a(5) = 129 = termwise products of (1, 1, 4, 13, 41) and (15, 10, 6, 3, 1) = (15 + 10 + 24 + 39 + 41). - Gary W. Adamson, Apr 10 2009
a(n) is the number of generalized compositions of n when there are i^2/2+i/2 different types of i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Dedrickson (Section 4.2) gives a bijection between colored compositions of n, where each part k has one of binomial(k+1,2) colors, and 0,1,2,3 strings of length n-1 avoiding 10, 20 and 21. Cf. A095263. For a refinement of this sequence counting binomial(k+1,2)-colored compositions by the number of parts see A127893. - Peter Bala, Sep 17 2013
|
|
REFERENCES
|
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 80.
|
|
LINKS
|
|
|
FORMULA
|
a(n) = Sum_{a=0..n} (Sum_{b=0..n} (Sum_{c=0..n} C(n-b-c,a)*C(n-a-c,b)*C(n-a-b,c))).
G.f.: (1 - x)^3/(1 - 4*x + 3*x^2 - x^3).
a(n) = 4*a(n-1) - 3*a(n-2) + a(n-3) for n>=4.
a(n) = Sum_{alpha = RootOf(-1+4*x-3*x^2+x^3)} (1/31)*(6 - 5*alpha - 3*alpha^2) * alpha^(-1-n).
For n>0, a(n) = Sum_{k=0..n-1} Sum_{i=0..k} Sum_{j=0..i} a(j). - Benoit Cloitre, Jan 26 2003
If p[i]=i(i+1)/2 and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det A. - Milan Janjic, May 02 2010
Recurrence equation: a(n) = Sum_{k = 1..n} 1/2*k*(k+1)*a(n-k) with a(0) = 1. - Peter Bala, Sep 19 2013
|
|
MAPLE
|
spec := [S, {S=Sequence(Prod(Z, Sequence(Z), Sequence(Z), Sequence(Z)))}, unlabeled]: seq(combstruct[count](spec, size=n), n=0..20);
f:= gfun:-rectoproc({a(n+4)-4*a(n+3)+3*a(n+2)-a(n+1), a(0) = 1, a(1) = 1, a(2) = 4, a(3) = 13}, a(n), `remember`):
|
|
MATHEMATICA
|
CoefficientList[Series[(-1+x)^3/(-1+4*x-3*x^2+x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 22 2012 *)
LinearRecurrence[{4, -3, 1}, {1, 1, 4, 13}, 30] (* Harvey P. Dale, Oct 04 2015 *)
|
|
PROG
|
(Magma) I:=[1, 1, 4, 13, 41, 129]; [n le 6 select I[n] else 4*Self(n-1) -3*Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 22 2012
(PARI) my(x='x+O('x^30)); Vec((1-x)^3/(1-4*x+3*x^2-x^3)) \\ G. C. Greubel, May 12 2019
(Sage) ((1-x)^3/(1-4*x+3*x^2-x^3)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 12 2019
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
encyclopedia(AT)pommard.inria.fr, Jan 25 2000
|
|
STATUS
|
approved
|
|
|
|