OFFSET
0,1
COMMENTS
The exact values are conjectured to be close to integers, but there is no sound basis for it as yet. The coincidence for the first 10 instances is however intriguing.
FORMULA
Sum[k^n/((2k)!/k!/(k+1)!), {k, \[Infinity]}]==Sum[Hypergeometric2F1[m+1, m+2, m+1/2, 1/4]StirlingS2[n, m]/(2m-1)!!/2^m(m+1)!m!, {m, 1, n}]; see Formula. Hypergeometric2F1[m, m+1, m-1/2, 1/4] equal to h[a_]:=h[a]=Apart[(4*(-3+2*a)*((-5+2*a)*h[ -2+a]-(-4+a)*h[ -1+a]))/(3*(-1+a)*a)]; h[1]:=2+(4*Pi)/(9*Sqrt[3]); h[0]:=1;
EXAMPLE
n float(n) Exact(n)
0 2.806133 2 + (4*Pi)/(9*Sqrt[3])
1 3.074844 2 + (16*Pi)/(27*Sqrt[3])
2 6.995495 (2*(567 + 52*Sqrt[3]*Pi))/243
3 20.986486 14 + (104*Pi)/(27*Sqrt[3])
4 79.000346 158/3 + (392*Pi)/(27*Sqrt[3])
MATHEMATICA
Hypergeometric2F1[a, a+1, a-1/2, 1/4] equals h[a_]:=h[a]=Apart[(4*(-3+2*a)*((-5+2*a)*h[ -2+a]-(-4+a)*h[ -1+a]))/(3*(-1+a)*a)]; h[1]:=2+(4*Pi)/(9*Sqrt[3]); h[0]:=1; Table[Round[Sum[h[m+1] StirlingS2[n, m]/(2m-1)!!/2^m (m+1)!m!, {m, 0, n}]], {n, 14}]
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Wouter Meeussen, Dec 05 2004
STATUS
approved