0,1

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

For Re(s) > 1, zeta(s) = lim_{x->oo} x / ( x - Sum_{p} floor(x/p^s) + Sum_{p<q} floor(x/(p*q)^s) - Sum_{p<q<r} floor(x/(p*q*r)^s) + ... ), where the sums are over distinct primes p, q, r..(inclusion-exclusion expansion of the Euler product). For even positive integers s = 2n, this gives the known rational number zeta(2n)/Pi^(2n), whose denominator is a(n). - Md. Rad Sarar Anando, Feb 09 2026

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).

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.

Noam D. Elkies, On the sums Sum((4k+1)^(-n),k,-inf,+inf), arXiv:math/0101168 [math.CA], 2001-2003.

Noam 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.

Andres Robles-Navarro, Shaun Cooper, Andreas A. Buchheit, Jonathan Busse, Antony Burrows, Odile Smits, and Peter Schwerdtfeger, Exact lattice summations for Lennard-Jones potentials coupled to a three-body Axilrod-Teller-Muto term applied to cuboidal phase transitions, arXiv:2504.07338 [cond-mat.mtrl-sci], 2025. See p. 36 (Appendix A).

Jonathan Sondow and Eric W. Weisstein, MathWorld: Riemann Zeta Function

Iickho Song, A recursive formula for even order harmonic series, J. Computational and Appl. Math., 21 (1988), 251-256.

Iickho 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.

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

(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

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

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

Cf. A046988 (numerators), A006003.

Sequence in context: A390164 A179214 A128265 * A087277 A177861 A218151

Adjacent sequences: A002429 A002430 A002431 * A002433 A002434 A002435

nonn,nice,easy,frac

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

approved