A086672 Stirling1 transform of Catalan numbers: Sum_{k=0..n} Stirling1(n,k)*binomial(2*k,k)/(k+1).
1, 1, 1, 1, 0, 1, -5, 29, -196, 1518, -13266, 129163, -1386572, 16270671, -207195495, 2845705719, -41930575740, 659781404944, -11041824881696, 195839234324062, -3669384701403344, 72423881548363354, -1501924519315744146, 32649768696532126439, -742432111781693213350
Offset: 0
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..200
Programs
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Mathematica
Table[Sum[StirlingS1[n, k] * CatalanNumber[k], {k, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Aug 04 2021 *) -
PARI
a(n)={sum(k=0, n, stirling(n,k,1) * binomial(2*k, k) / (k+1))} \\ Andrew Howroyd, Jan 27 2020
Formula
E.g.f.: hypergeom([1/2], [2], 4*log(1+x)) = (1+x)^2*(BesselI(0, 2*log(1+x))-BesselI(1, 2*log(1+x))).
Let C(m) be the m-th Catalan number, A000108(m). Let S(m, n) = an unsigned Stirling number of the first kind. Then a(m) = sum{k=0 to m} S(m, k) C(k) (-1)^(k+m). - Leroy Quet, Jan 23 2004
E.g.f. f(x) satisfies f(x) = 1 + integral{0 to x} f(y) f((x-y)/(1+y))/(1+y) dy. - Leroy Quet, Jan 25 2004
For n > 1, a(n) = (A201950(n+1) - (3*n-2)*A201950(n) + n*(3*n-7)*A201950(n-1) - (n-4)*(n-1)*n*A201950(n-2)) * (-1)^n/2. - Vaclav Kotesovec, May 04 2024
Extensions
Terms a(21) and beyond from Andrew Howroyd, Jan 27 2020
Comments