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%I #15 May 29 2017 03:17:07
%S 6,90,945,9450,93555,924041,9121612,90030844,888579011,8769948429,
%T 86555983552,854273468992,8431341566236,83214006759229,
%U 821289329637860,8105800788023426,80001047145799660,789578687036411293
%N Floor of Pi^(2*n)/Zeta(2*n).
%H Vincenzo Librandi, <a href="/A100594/b100594.txt">Table of n, a(n) for n = 1..200</a>
%e a(1)=6 because Zeta(2*1)=Pi^2/6 implies Pi^2/Zeta(2)=6 and floor(6)=6.
%e a(6)=924041 because Zeta(2*6)=691/638512875*Pi^12 implies Pi^12/Zeta(12)=638512875/691 and floor(638512875/691)=924041.
%p seq(simplify(floor(Pi^(2*k)/Zeta(2*k))),k=1..24);
%t Table[Floor[Pi^(2*n)/Zeta[2*n]],{n,20}] (* _Terry D. Grant_, May 28 2017 *)
%o (PARI) {a(n)=if(n<1, 0, floor(-2*(2*n)!/(-4)^n/bernfrac(2*n)))} /* _Michael Somos_, Feb 18 2007 */
%Y Cf. A002432, A046988.
%K nonn
%O 1,1
%A Joseph Biberstine (jrbibers(AT)indiana.edu), Nov 30 2004