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%I M4283 N1790 #175 Mar 18 2023 08:49:14
%S 2,6,90,945,9450,93555,638512875,18243225,325641566250,38979295480125,
%T 1531329465290625,13447856940643125,201919571963756521875,
%U 11094481976030578125,564653660170076273671875,5660878804669082674070015625,62490220571022341207266406250
%N Denominators of zeta(2*n)/Pi^(2*n).
%C Also denominators in expansion of Psi(x).
%C For n >= 1 a(n) is always divisible by 3 (by the von Staudt-Clausen theorem, see A002445).
%C 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
%C 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
%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
%D 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.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H T. D. Noe and J.P. Martin-Flatin, <a href="/A002432/b002432.txt">Table of n, a(n) for n = 0..250</a> (first 100 terms were computed by T. D. Noe).
%H G. Campbell, <a href="https://www.linkedin.com/groups/Some-series-like-ζ-3-4510047.S.5985755984868421632">Some series like ζ(3), ζ(5), ζ(7).</a> Number Theory group on LinkedIn.com, March 2015.
%H N. D. Elkies, <a href="https://arxiv.org/abs/math/0101168">On the sums Sum((4k+1)^(-n),k,-inf,+inf)</a>, arXiv:math/0101168 [math.CA], 2001-2003.
%H N. D. Elkies, <a href="http://www.jstor.org/stable/3647742">On the sums Sum_{k = -infinity .. infinity} (4k+1)^(-n)</a>, Amer. Math. Monthly, 110 (No. 7, 2003), 561-573.
%H Masato Kobayashi and Shunji Sasaki, <a href="https://arxiv.org/abs/2202.11835">Values of zeta-one functions at positive even integers</a>, arXiv:2202.11835 [math.NT], 2022. See p. 4.
%H J. Sondow and E. W. Weisstein, <a href="http://mathworld.wolfram.com/RiemannZetaFunction.html">MathWorld: Riemann Zeta Function</a>
%H I. Song, <a href="http://dx.doi.org/10.1016/0377-0427(88)90274-9">A recursive formula for even order harmonic series</a>, J. Computational and Appl. Math., 21 (1988), 251-256.
%H I. Song, <a href="/A002432/a002432.pdf"> A recursive formula for even order harmonic series</a>, J. Computational and Appl. Math., 21 (1988), 251-256. [Annotated scanned copy]
%H <a href="/index/Z#zeta_function">Index entries for zeta function</a>.
%F 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
%F zeta(2n)/(2i * ( log(1-i)-log(1+i) ))^(2n) = zeta(2n)/(-i*log(-1))^(2n). - _Eric Desbiaux_, Dec 12 2008
%F zeta(2n) = Sum_{k >= 1} k^(-2n) = (-1)^(n-1)*B_{2n}*2^(2n-1)*Pi^(2n)/(2n)!.
%F a(n) = -A046988(n)*A010050(n)*A002445(n)/(A009117(n)*A000367(n))
%F a(n) = sqrt(denominator(Sum_{i>=1} A000005(i)/i^2n)). - _Enrique Pérez Herrero_, Jan 19 2012
%F Sum_{k >= 1} zeta(2k)*x^(2k) = (1-Pi*x*cot(Pi*x))/2. - _Chris Boyd_, Dec 21 2015
%F a(n) = denominator([x^(2*n)] -x*cot(x)/2). - _Peter Luschny_, Jun 07 2020
%e (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, ...
%e 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 + ...
%e a(5) = 93555 = 10!/(2^9 * B(10)) = 3628800/(512*5/66). - _Frank Ellermann_, Apr 03 2020
%p seq(denom(Zeta(2*n)/Pi^(2*n)),n=0..24); # _Martin Renner_, Sep 07 2016
%p A002432List := proc(len) series(-x*cot(x)/2, x, 2*len+1):
%p seq(denom(coeff(%, x, n)), n=0..2*len-1, 2) end:
%p A002432List(17); # _Peter Luschny_, Jun 07 2020
%t Table[Denominator[Zeta[2 n]/Pi^(2 n)], {n, 0, 30}] (* _Artur Jasinski_, Mar 11 2010 *)
%t Denominator[Zeta[2*Range[0, 20]]] (* _Harvey P. Dale_, Sep 05 2013 *)
%o (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
%o (PARI) a002432(n) = denominator(polcoeff((1-x*cotan(x))/2,n*2))
%o default(seriesprecision, 33); for(i=0,16,print1(a002432(i),",")) \\ _Chris Boyd_, Dec 21 2015
%Y Cf. A046988 (numerators), A006003.
%K nonn,nice,easy,frac
%O 0,1
%A _N. J. A. Sloane_
%E Formula and link from _Henry Bottomley_, Mar 10 2000
%E Formula corrected by Bjoern Boettcher, May 15 2003
%E Corrected and edited by _M. F. Hasler_, Mar 29 2015
%E a(0) = 2 prepended by _Peter Luschny_, Jun 07 2020
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