%I #33 Dec 08 2021 05:56:35
%S 0,0,0,1,2,3,4,6,7,9,11,13,16,19,22,25,28,31,35,39,43,47,52,57,62,67,
%T 72,78,83,89,95,102,108,115,122,129,136,144,152,160,168,176,185,193,
%U 202,212,221,231,240,250,260,271,281,292,303,314,326,337,349,361,373
%N a(n) = floor(abs(B(2*n + 2)/B(2*n))) where B(n) is the n-th Bernoulli number.
%D Glaisher, J. W. L.; Tables of the first 250 Bernoulli numbers. Trans. Cambridge Phil. Soc. 12 (1873), 384-391.
%D Peters, J. and Stein, J., Matematische Tafeln. Revised Russian Edition, 1968, Moscow.
%H Robert Israel, <a href="/A047872/b047872.txt">Table of n, a(n) for n = 0..10000</a>
%H Maths StackExchange, <a href="https://math.stackexchange.com/questions/3181047/bernoulli-numbers-and-pi2">Bernoulli numbers and Pi^2</a>, 2019.
%H <a href="/index/Be#Bernoulli">Index entries for sequences related to Bernoulli numbers</a>.
%F a(n) = floor( (n+1)*(2*n+1)/(2*Pi^2)) (conjectured). - _Bill McEachen_, Dec 08 2021
%F A002939(n+1)*B(2*n)/B(2*(n+1)) = -(2*Pi)^2*(1 + O(1/4^n)). See the StackExchange link. - _Peter Luschny_, Dec 08 2021
%e a(3) = floor(abs(B(4)/B(3))) = floor((1/30)/(1/42)) = floor(7/5) = floor(1.4) = 1.
%e a(249) = floor(abs(B(250)/B(249))) = 6319.
%p seq(floor(abs(bernoulli(2*n+2)/bernoulli(2*n))),n=0..200); # _Robert Israel_, Jun 27 2018
%t Table[Floor[Abs[BernoulliB[2*n + 2]/BernoulliB[2*n]]], {n, 0, 60}] (* _T. D. Noe_, Jun 27 2013 *)
%o (PARI) a(n) = floor(abs(bernfrac(2*n+2)/bernfrac(2*n))) \\ _Michel Marcus_, Jun 27 2013
%Y Cf. A034972, A034971, A002939.
%K easy,nonn
%O 0,5
%A _Labos Elemer_