A002432 Denominators of zeta(2*n)/Pi^(2*n).

2, 6, 90, 945, 9450, 93555, 638512875, 18243225, 325641566250, 38979295480125, 1531329465290625, 13447856940643125, 201919571963756521875, 11094481976030578125, 564653660170076273671875, 5660878804669082674070015625, 62490220571022341207266406250

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• nice
• easy
• frac

Comments

Also denominators in expansion of Psi(x).

For n >= 1 a(n) is always divisible by 3 (by the von Staudt-Clausen theorem, see A002445).

A comment due to G. Campbell: The original approach taken by Euler was to derive the infinite product for sin(Pi*x)/(Pi*x) equal to (1 - x^2/1^2) (1 - x^2/2^2)(1 - x^2/3^2) ... treating sin(Pi*x)/(Pi*x) as if it were a polynomial. Differentiating the logarithm of both sides and equating coefficients gives all of the zeta function values for 2, 4, 6, 8, .... - M. F. Hasler, Mar 29 2015

Note that 2n+1 divides a(n) for every n. If 2n+1 > 9 is composite, then (2n+1)^2 divides a(n). If 2n+1 is prime, then (2n+1)^2 does not divide a(n). My theorem: for n > 4, (2n+1)^2 divides a(n) if and only if the number 2n+1 is composite. - Thomas Ordowski, Nov 07 2022

Examples

			(zeta(2n)/Pi^(2n), n = 0, 1, 2, 3, ...) = (-1/2, 1/6, 1/90, 1/945, 1/9450, 1/93555, 691/638512875, 2/18243225, 3617/325641566250, ...), i.e.: zeta(0) = -1/2, zeta(2) = Pi^2/6, zeta(4) = Pi^4/90, zeta(6) = Pi^6/945, zeta(8) = Pi^8/9450, zeta(10) = Pi^10/93555, zeta(12) = Pi^12*691/638512875, ...
In Maple, series(Psi(x),x,20) gives -1*x^(-1) + (-gamma) + 1/6*Pi^2*x + (-Zeta(3))*x^2 + 1/90*Pi^4*x^3 + (-Zeta(5))*x^4 + 1/945*Pi^6*x^5 + (-Zeta(7))*x^6 + 1/9450*Pi^8*x^7 + (-Zeta(9))*x^8 + 1/93555*Pi^10*x^9 + ...
a(5) = 93555 = 10!/(2^9 * B(10)) = 3628800/(512*5/66). - _Frank Ellermann_, Apr 03 2020
		

References

• L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
• 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. 84.
• 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

• T. D. Noe and J.P. Martin-Flatin, Table of n, a(n) for n = 0..250 (first 100 terms were computed by T. D. Noe).
• G. Campbell, Some series like ζ(3), ζ(5), ζ(7). Number Theory group on LinkedIn.com, March 2015.
• N. D. Elkies, On the sums Sum((4k+1)^(-n),k,-inf,+inf), arXiv:math/0101168 [math.CA], 2001-2003.
• N. D. Elkies, On the sums Sum_{k = -infinity .. infinity} (4k+1)^(-n), Amer. Math. Monthly, 110 (No. 7, 2003), 561-573.
• Masato Kobayashi and Shunji Sasaki, Values of zeta-one functions at positive even integers, arXiv:2202.11835 [math.NT], 2022. See p. 4.
• J. Sondow and E. W. Weisstein, MathWorld: Riemann Zeta Function
• I. Song, A recursive formula for even order harmonic series, J. Computational and Appl. Math., 21 (1988), 251-256.
• I. Song, A recursive formula for even order harmonic series, J. Computational and Appl. Math., 21 (1988), 251-256. [Annotated scanned copy]
• Index entries for zeta function.

Crossrefs

Cf. A046988 (numerators), A006003.

Programs

Maple

seq(denom(Zeta(2*n)/Pi^(2*n)),n=0..24); # Martin Renner, Sep 07 2016
A002432List := proc(len) series(-x*cot(x)/2, x, 2*len+1):
seq(denom(coeff(%, x, n)), n=0..2*len-1, 2) end:
A002432List(17); # Peter Luschny, Jun 07 2020

Mathematica

Table[Denominator[Zeta[2 n]/Pi^(2 n)], {n, 0, 30}] (* Artur Jasinski, Mar 11 2010 *)
Denominator[Zeta[2*Range[0, 20]]] (* Harvey P. Dale, Sep 05 2013 *)

PARI

a(n)=numerator(bestappr(Pi^(2*n)/zeta(2*n))) \\ Requires sufficient realprecision. The standard value of 38 digits yields erroneous values for n>9. \p99 is more than enough to get the 3 lines of displayed data. - M. F. Hasler, Mar 29 2015

PARI

a002432(n) = denominator(polcoeff((1-x*cotan(x))/2,n*2))
default(seriesprecision, 33); for(i=0,16,print1(a002432(i),",")) \\ Chris Boyd, Dec 21 2015

Formula

Sum_{n>=1} 2/(n^2 + 1) = Pi*coth(Pi)-1. 2*Sum_{k>=1} (-1)^(k + 1)/n^(2*k) = 2/(n^2+1). - Shmuel Spiegel (shmualm(AT)hotmail.com), Aug 13 2001

zeta(2n)/(2i * ( log(1-i)-log(1+i) ))^(2n) = zeta(2n)/(-i*log(-1))^(2n). - Eric Desbiaux, Dec 12 2008

zeta(2n) = Sum_{k >= 1} k^(-2n) = (-1)^(n-1)*B_{2n}*2^(2n-1)*Pi^(2n)/(2n)!.

a(n) = -A046988(n)*A010050(n)*A002445(n)/(A009117(n)*A000367(n))

a(n) = sqrt(denominator(Sum_{i>=1} A000005(i)/i^2n)). - Enrique Pérez Herrero, Jan 19 2012

Sum_{k >= 1} zeta(2k)*x^(2k) = (1-Pi*x*cot(Pi*x))/2. - Chris Boyd, Dec 21 2015

a(n) = denominator([x^(2*n)] -x*cot(x)/2). - Peter Luschny, Jun 07 2020

Extensions

Formula and link from Henry Bottomley, Mar 10 2000

Formula corrected by Bjoern Boettcher, May 15 2003

Corrected and edited by M. F. Hasler, Mar 29 2015

a(0) = 2 prepended by Peter Luschny, Jun 07 2020