A100666 Sum_{k>=1} k^n/Catalan(k) rounded to nearest integer.

3, 3, 7, 21, 79, 357, 1879, 11277, 75967, 567381, 4652071, 41534493, 401057935, 4164175845, 46260731384, 547489559470, 6876483788377, 91352567576937, 1279774932585453, 18855298837939164, 291449116254193528

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.

Links

Table of n, a(n) for n=0..20.

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

Sequence in context: A097334 A214496 A046631 * A262375 A232368 A333924

Adjacent sequences: A100663 A100664 A100665 * A100667 A100668 A100669

Keyword

easy,nonn

Author

Wouter Meeussen, Dec 05 2004

Status

approved