OFFSET
0,1
COMMENTS
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
LINKS
Harvey P. Dale, Table of n, a(n) for n = 0..201
Jean-Claude Babois, Personal communication to N. J. A. Sloane, Apr 29 2015 (Pages 4 and 1, Pages 2 and 3)
FORMULA
The ratio of successive terms is 8*A000384(n).
a(n) = 2^(2*n+2)*(2*n)!. - Vaclav Kotesovec, May 05 2015
EXAMPLE
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).
MAPLE
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)];
MATHEMATICA
lst={4}; Do[AppendTo[lst, 8*n*(2*n-1)*Last[lst]], {n, 1, 12}]; lst (* Ivan N. Ianakiev, May 04 2015 *)
Table[2^(2*n+2)*(2*n)!, {n, 0, 15}] (* Vaclav Kotesovec, May 05 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 *)
PROG
(Magma) [2^(2*n+2)*Factorial(2*n): n in [0..15]]; // Vincenzo Librandi, May 23 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, May 04 2015
STATUS
approved