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A257583
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a(0)=4; thereafter a(n)=8*n*(2*n-1)*a(n-1).
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1
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4, 32, 1536, 184320, 41287680, 14863564800, 7847962214400, 5713316492083200, 5484783832399872000, 6713375410857443328000, 10204330624503313858560000, 18857602994082124010618880000, 41637587410933329815446487040000
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OFFSET
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0,1
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COMMENTS
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It appears that for n>=1,
Sum_{k >= 0} ( 1/(4k+1)^(2n+1) - 1/(4k+3)^(2n+1) ) = E_n*Pi^(2n+1)/a(n),
where E_n = A000364(n) are the Euler numbers. For n=0, one must subtract 2/3 from the right-hand side.
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
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LINKS
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FORMULA
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The ratio of successive terms is 8*A000384(n).
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EXAMPLE
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For n=0, Sum_{k >= 0} ( 1/(4k+1)^(2n+1) - 1/(4k+3)^(2n+1) )
= 2*sum(1/((4*n+1)*(4*n+3)),n=1..infinity) = Pi/4 - 2/3 = E_0*Pi/a(0) - 2/3.
For n=1, the sum is Pi^3/32 = E_1*Pi^3/a(1).
For n=2, the sum is 5*Pi^5/1536 = E_2*Pi^5/a(2).
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MAPLE
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f:=proc(n) option remember; if n=0 then 4 else 8*n*(2*n-1)*f(n-1); fi; end;
[seq(f(n), n=0..20)];
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MATHEMATICA
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lst={4}; Do[AppendTo[lst, 8*n*(2*n-1)*Last[lst]], {n, 1, 12}]; lst (* Ivan N. Ianakiev, May 04 2015 *)
f[x_] := x Log[x] + (1 - x) Log[1 - x];
Table[D[f[x], {x, 2n}] /. x -> 1/2, {n, 1, 14}](* Paul Reiser, May 23 2015*)
nxt[{n_, a_}]:={n+1, 8a(1+n)(1+2n)}; NestList[nxt, {0, 4}, 20][[All, 2]] (* Harvey P. Dale, Jan 05 2023 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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