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A100666
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Sum_{k>=1} k^n/Catalan(k) rounded to nearest integer.
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0
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3, 3, 7, 21, 79, 357, 1879, 11277, 75967, 567381, 4652071, 41534493, 401057935, 4164175845, 46260731384, 547489559470, 6876483788377, 91352567576937, 1279774932585453, 18855298837939164, 291449116254193528
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OFFSET
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0,1
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COMMENTS
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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.
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LINKS
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FORMULA
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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;
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EXAMPLE
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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])
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MATHEMATICA
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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}]
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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