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

0,7

Comments

1, 1, 1, 0, 1, -5, 29, -196, ... is the Stirling1 transform of the Motzkin numbers A001006. - Philippe Deléham, May 27 2015

Links

Andrew Howroyd, Table of n, a(n) for n = 0..200

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

a(n) = Sum_{k = 0..n} A048994(n, k) * A000108(k). - Philippe Deléham, May 27 2015

a(n+1) = Sum_{k = 0..n} A048994(n,k) * A001006(k). - Philippe Deléham, May 27 2015

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

Mathematica

Table[Sum[StirlingS1[n, k] * CatalanNumber[k], {k, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Aug 04 2021 *)

Prog

(PARI) a(n)={sum(k=0, n, stirling(n, k, 1) * binomial(2*k, k) / (k+1))} \\ Andrew Howroyd, Jan 27 2020

Crossrefs

Cf. A000108, A001006, A008275, A048994, A064856.

Cf. A201950, A201952, A306335.

Sequence in context: A225030 A188143 A095000 * A324962 A306932 A201856

Adjacent sequences: A086669 A086670 A086671 * A086673 A086674 A086675

Keyword

easy,sign

Author

Vladeta Jovovic, Sep 12 2003

Extensions

Terms a(21) and beyond from Andrew Howroyd, Jan 27 2020

Status

approved