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A002425
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Denominator of Pi^(2n)/(Gamma(2n)*(1-2^(-2n))*zeta(2n)).
(Formerly M5036 N2174)
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29
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1, 1, 1, 17, 31, 691, 5461, 929569, 3202291, 221930581, 4722116521, 968383680827, 14717667114151, 2093660879252671, 86125672563201181, 129848163681107301953, 868320396104950823611, 209390615747646519456961
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OFFSET
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1,4
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COMMENTS
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Differs from the absolute values of A275994 the first time at index 60.
Consider the C(k)-summation process for divergent series: the series Sum((-1)^n*(n+1)^k) == 1 - 2^k + 3^k - 4^k + ..., summable C(1) to the value 1/2 for k = 0, is for each k >= 1 exactly summable C(k+1) to the sum s(k+1) = (2^(k+1)-1)*B(k+1)/ (k+1) and so a(n) = abs(numerator(s(2n))). - Benoit Cloitre, Apr 27 2002
a(n) is the absolute value of the constant term of the Euler polynomial E_{2n-1} times the even part of 2n. - Peter Luschny, Nov 26 2010
Let E_m(x) = x^m + Sum_{odd k=1..m} e_k(m)*x^(m-k) be the Euler polynomial, let 2*n-1 <= m. Show that the expression c(m,n) = |e_(2*n-1)(m)|/binomial(m,2*n-1) does not depend on m and c(m,n) = a(n)/A006519(2*n). Indeed, by the formula in the Shevelev link |e_(2*n-1)(m)| = binomial(m,2*n-1)*(4^n-1)*B_(2*n)/n. On the other hand, by Cloitre's formula, we have a(n) = (4^n-1)*|B_(2*n)|*2^A001511(n) /n. Taking into account that 2^A001511 = A006519(2*n) we obtain the claimed equality. Since sign(e_k(n)) = (-1)^((k+1)/2), we have the following application of the sequence: e_k(n) = (-1)^((k+1)/2))*a((k+1)/2)*binomial(n,k)/A006519(k+1). (End)
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REFERENCES
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A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 73.
S. A. Joffe, Sums of like powers of natural numbers, Quart. J. Pure Appl. Math. 46 (1914), 33-51.
Konrad Knopp, Theory and application of infinite series, Divergent series, Dover, p. 479
L. Oettinger, Archiv. Math. Phys., 26 (1856), see esp. p. 5.
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) = (-1)^n/n*(1 - 4^n)*B(2*n)*2^A001511(n) where B(k) denotes the k-th Bernoulli number. - Benoit Cloitre, Dec 30 2003
This is different from the sequence of numerators of the expansion of cosec(x) - cot(x) - see A089171.
a(n) = denominator(4*n/((2^(2*n)-1)*bernoulli(2*n))).
E.g.f.: a(n) = numerator((2*n+1)!*[x^(2*n+1)](1/(1+1/exp(x)))). - Peter Luschny, Jul 12 2012
For every positive integers n,k we have a(n) = (-1)^(n+k)*N(2*n-1,k) + 2*(-1)^(n-1)*A006519(2*n)*(1^(2*n-1)-2^(2*n-1)+..+(-1)^k*(k-1)^(2*n-1)), where N(n,k) is the numerator of Euler(n,k). So, the right hand side is an invariant of k. - Vladimir Shevelev, Sep 19 2017
a(n) = numerator(r(n)) where r(n) = (-1)^binomial(2*n, 2)*Sum_{k=1..2*n}(-1)^k*Stirling2(2*n, k)*2^(-k)*(k-1)!. - Peter Luschny, May 24 2020
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MAPLE
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A002425_list := proc(n) 1/(1+1/exp(z)); series(%, z, 2*n+4);
seq(numer((-1)^i*(2*i+1)!*coeff(%, z, 2*i+1)), i=0..n) end;
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MATHEMATICA
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a[n_]:= If[n<1, 0, With[{m = 2n-1}, Numerator[ m! SeriesCoefficient[ Tan[x/2], {x, 0, m}]]]] (* Michael Somos, Sep 14 2013 *)
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PROG
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(PARI) for(n=1, 20, print1(abs(numerator(2*bernfrac(2*n)*(4^n-1)/(2*n))), ", "))
(PARI) a(n)=if(n<1, 0, (-1)^n/n*(1-4^n)*bernfrac(2*n)*2^valuation(2*n, 2))
(PARI) a(n)=(-1)^n*4*bitand(n, -n)*polylog(1-2*n, -1); \\ Peter Luschny, Nov 22 2012
(Sage)
T = [0]*n; T[0] = 1; S = [0]*n; k2 = 0
for k in (1..n-1): T[k] = k*T[k-1]
for k in (1..n):
if is_odd(k): S[k-1] = 4*k2; k2 += 1
else: S[k-1] = S[k2-1]+2*k2-1
for j in (k..n-1): T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
return [T[j]>>S[j] for j in (0..n-1)]
(Sage) [denominator(4*n/((4^n-1)*bernoulli(2*n))) for n in (1..20)] # G. C. Greubel, Jul 03 2019
(Magma) [Denominator(4*n/((4^n-1)*Bernoulli(2*n))): n in [1..20]]; // G. C. Greubel, Jul 03 2019
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CROSSREFS
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KEYWORD
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nonn,frac,easy
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AUTHOR
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EXTENSIONS
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The n=15 term was formerly incorrectly given as 86125672563301143.
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STATUS
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approved
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