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A211611
a(n) = Sum_{k=1..n-1} C(k)^n, where C(k) is a Catalan number.
3
1, 9, 642, 540982, 5496576970, 698491214560174, 1147342896257677900291, 25005346993500437111980892595, 7381619397278667883874693730628586499, 30009934325456999669083059570156145437948880627, 1703283943023520710008632777768663744247664926649672215939
OFFSET
2,2
COMMENTS
The C(k) are the Catalan numbers, C(k) = A000108(k) = (2k)!/(k!*(k+1)!) = C(2*k,k)/(k+1).
p divides a(p) for prime p of the form p = 6k + 1 (A002476).
LINKS
Eric Weisstein's World of Mathematics, Catalan Number
FORMULA
a(n) = Sum_{k=1..n-1} binomial(2*k, k)/(k+1)^n.
a(n) ~ exp(3/8) * 4^(n^2-n) / (Pi^(n/2) * n^(3*n/2)). - Vaclav Kotesovec, Mar 03 2014
MATHEMATICA
Table[ Sum[ (Binomial[2 k, k]/(k + 1))^n, {k, 1, n - 1}], {n, 2, 13}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Alexander Adamchuk, Apr 17 2012
STATUS
approved