A100666 - OEIS

%I #8 Jan 31 2013 12:03:12

%S 3,3,7,21,79,357,1879,11277,75967,567381,4652071,41534493,401057935,

%T 4164175845,46260731384,547489559470,6876483788377,91352567576937,

%U 1279774932585453,18855298837939164,291449116254193528

%N Sum_{k>=1} k^n/Catalan(k) rounded to nearest integer.

%C 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.

%F 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;

%e n float(n) Exact(n)

%e 0 2.806133 2 + (4*Pi)/(9*Sqrt[3])

%e 1 3.074844 2 + (16*Pi)/(27*Sqrt[3])

%e 2 6.995495 (2*(567 + 52*Sqrt[3]*Pi))/243

%e 3 20.986486 14 + (104*Pi)/(27*Sqrt[3])

%e 4 79.000346 158/3 + (392*Pi)/(27*Sqrt[3])

%t 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}]

%K easy,nonn

%O 0,1

%A _Wouter Meeussen_, Dec 05 2004