A002425 Denominator of Pi^(2n)/(Gamma(2n)*(1-2^(-2n))*zeta(2n)). (Formerly M5036 N2174)

1, 1, 1, 17, 31, 691, 5461, 929569, 3202291, 221930581, 4722116521, 968383680827, 14717667114151, 2093660879252671, 86125672563201181, 129848163681107301953, 868320396104950823611, 209390615747646519456961

Offset

1,4

Comments

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

Odd part of tangent numbers A000182 (even part is 2^A101921(n)). - Ralf Stephan, Dec 21 2004

(-1)^n*a(n+1) is the numerator of Euler(2n+1,1). - N. J. A. Sloane, Nov 10 2009 (a misprint corrected by Vladimir Shevelev, Sep 18 2017)

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

From Vladimir Shevelev, Aug 31 2017: (Start)

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)

References

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).

Links

N. J. A. Sloane, Table of n, a(n) for n = 1..300

H. Cohn, Some elementary aspects of modular functions in several variables, Bull. Am. Math. Soc., Sept. 1965, 681ff, esp. p. 688.

Ren Guan, K_0 groups of noncommutative R^2n, arXiv:2208.06253 [math.RA], 2022. See p. 22.

S. A. Joffe, Sums of like powers of natural numbers, Quart. J. Pure Appl. Math. 46 (1914), 33-51. [Annotated scanned copy of pages 38-51 only, plus notes]

Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")

Vladimir Shevelev, On a Luschny question, arXiv:1708.08096 [math.NT], 2017.

Formula

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.

From Johannes W. Meijer, May 24 2009: (Start)

a(n) = denominator(4*n/((2^(2*n)-1)*bernoulli(2*n))).

Equals A160469(n)/A048896(n-1).

Equals A089171(n)*A089170(n-1). (End)

E.g.f.: a(n) = numerator((2*n+1)!*[x^(2*n+1)](1/(1+1/exp(x)))). - Peter Luschny, Jul 12 2012

a(n) = numerator(abs(2*(4^n-1)*zeta(1-2*n))). - Jean-François Alcover, Oct 16 2013

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

a(n) = 2*(-1)^n*A335956(2*n)*zeta(1-2*n). - Peter Luschny, Aug 30 2020

Maple

A002425 := n -> (-1)^n*euler(2*n-1, 0)*2^padic[ordp](2*n, 2); # Peter Luschny, Nov 26 2010
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;
A002425_list(17); # Peter Luschny, Jul 12 2012

Mathematica

a[n_]:= (-1)^(n-1) * Numerator[EulerE[2n-1, 1]]; Table[a[n], {n, 1, 20}] (* Jean-François Alcover, Sep 20 2011, after N. J. A. Sloane's comment *)
a[n_]:= If[n<1, 0, With[{m = 2n-1}, Numerator[ m! SeriesCoefficient[ Tan[x/2], {x, 0, m}]]]] (* Michael Somos, Sep 14 2013 *)
Table[2*(4^n-1)*Zeta[1-2n] // Abs // Numerator, {n, 1, 20}] (* Jean-François Alcover, Oct 16 2013 *)

Prog

(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)
def A002425_list(n):
    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)]
A002425_list(20)  # Peter Luschny, Nov 17 2012
(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

Crossrefs

Numerator given by A037239.

Different from A089171, A275994.

Cf. A048896, A089170, A089171, A160469, A335956.

Sequence in context: A089171 A279370 A276592 * A275994 A046990 A059212

Adjacent sequences: A002422 A002423 A002424 * A002426 A002427 A002428

Keyword

nonn,frac,easy

Author

N. J. A. Sloane

Extensions

The n=15 term was formerly incorrectly given as 86125672563301143.

Formula and cross-references edited by Johannes W. Meijer, May 21 2009

Status

approved