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%I #26 Aug 04 2021 10:10:35
%S 1,1,1,1,0,1,-5,29,-196,1518,-13266,129163,-1386572,16270671,
%T -207195495,2845705719,-41930575740,659781404944,-11041824881696,
%U 195839234324062,-3669384701403344,72423881548363354,-1501924519315744146,32649768696532126439,-742432111781693213350
%N Stirling1 transform of Catalan numbers: Sum_{k=0..n} Stirling1(n,k)*binomial(2*k,k)/(k+1).
%C 1, 1, 1, 0, 1, -5, 29, -196, ... is the Stirling1 transform of the Motzkin numbers A001006. - _Philippe Deléham_, May 27 2015
%H Andrew Howroyd, <a href="/A086672/b086672.txt">Table of n, a(n) for n = 0..200</a>
%F 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))).
%F 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
%F 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
%F a(n) = Sum_{k = 0..n} A048994(n, k) * A000108(k). - _Philippe Deléham_, May 27 2015
%F a(n+1) = Sum_{k = 0..n} A048994(n,k) * A001006(k). - _Philippe Deléham_, May 27 2015
%t Table[Sum[StirlingS1[n, k] * CatalanNumber[k], {k, 0, n}], {n, 0, 25}] (* _Vaclav Kotesovec_, Aug 04 2021 *)
%o (PARI) a(n)={sum(k=0, n, stirling(n,k,1) * binomial(2*k, k) / (k+1))} \\ _Andrew Howroyd_, Jan 27 2020
%Y Cf. A000108, A001006, A008275, A048994, A064856.
%K easy,sign
%O 0,7
%A _Vladeta Jovovic_, Sep 12 2003
%E Terms a(21) and beyond from _Andrew Howroyd_, Jan 27 2020
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