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%I #46 Jan 05 2023 15:38:18
%S 4,32,1536,184320,41287680,14863564800,7847962214400,5713316492083200,
%T 5484783832399872000,6713375410857443328000,
%U 10204330624503313858560000,18857602994082124010618880000,41637587410933329815446487040000
%N a(0)=4; thereafter a(n)=8*n*(2*n-1)*a(n-1).
%C It appears that for n>=1,
%C Sum_{k >= 0} ( 1/(4k+1)^(2n+1) - 1/(4k+3)^(2n+1) ) = E_n*Pi^(2n+1)/a(n),
%C where E_n = A000364(n) are the Euler numbers. For n=0, one must subtract 2/3 from the right-hand side.
%C It appears that the entries are the numerators of the 2n-th coefficients of the Taylor series expansion of f[x] = x Log[x]+(1-x)Log[1-x] about x=1/2, beginning at n=1, odd derivatives being zero, denominators being 2n!. In other words, the 2n-order derivative of f[x] evaluated at x=1/2 for n=1,2,3... (See Mathematica below). - _Paul Reiser_, May 23 2015
%H Harvey P. Dale, <a href="/A257583/b257583.txt">Table of n, a(n) for n = 0..201</a>
%H Jean-Claude Babois, Personal communication to N. J. A. Sloane, Apr 29 2015 (<a href="/A257583/a257583_1.png">Pages 4 and 1</a>, <a href="/A257583/a257583.png">Pages 2 and 3)</a>
%F The ratio of successive terms is 8*A000384(n).
%F a(n) = 2^(2*n+2)*(2*n)!. - _Vaclav Kotesovec_, May 05 2015
%e For n=0, Sum_{k >= 0} ( 1/(4k+1)^(2n+1) - 1/(4k+3)^(2n+1) )
%e = 2*sum(1/((4*n+1)*(4*n+3)),n=1..infinity) = Pi/4 - 2/3 = E_0*Pi/a(0) - 2/3.
%e For n=1, the sum is Pi^3/32 = E_1*Pi^3/a(1).
%e For n=2, the sum is 5*Pi^5/1536 = E_2*Pi^5/a(2).
%p f:=proc(n) option remember; if n=0 then 4 else 8*n*(2*n-1)*f(n-1); fi; end;
%p [seq(f(n),n=0..20)];
%t lst={4};Do[AppendTo[lst,8*n*(2*n-1)*Last[lst]],{n,1,12}];lst (* _Ivan N. Ianakiev_, May 04 2015 *)
%t Table[2^(2*n+2)*(2*n)!,{n,0,15}] (* _Vaclav Kotesovec_, May 05 2015 *)
%t f[x_] := x Log[x] + (1 - x) Log[1 - x];
%t Table[D[f[x], {x, 2n}] /. x -> 1/2, {n, 1, 14}](* _Paul Reiser_, May 23 2015*)
%t nxt[{n_,a_}]:={n+1,8a(1+n)(1+2n)}; NestList[nxt,{0,4},20][[All,2]] (* _Harvey P. Dale_, Jan 05 2023 *)
%o (Magma) [2^(2*n+2)*Factorial(2*n): n in [0..15]]; // _Vincenzo Librandi_, May 23 2015
%Y Cf. A000364, A000384.
%K nonn
%O 0,1
%A _N. J. A. Sloane_, May 04 2015
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