A194589 a(n) = A194588(n) - A005043(n); complementary Riordan numbers.
0, 0, 1, 1, 5, 11, 34, 92, 265, 751, 2156, 6194, 17874, 51702, 149941, 435749, 1268761, 3700391, 10808548, 31613474, 92577784, 271407896, 796484503, 2339561795, 6877992334, 20236257626, 59581937299, 175546527727, 517538571125, 1526679067331, 4505996000730
Offset: 0
Keywords
Links
- Peter Luschny, The lost Catalan numbers.
Programs
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Maple
# First method, describes the derivation: A056040 := n -> n!/iquo(n,2)!^2: A057977 := n -> A056040(n)/(iquo(n,2)+1); A001006 := n -> add(binomial(n,k)*A057977(k)*irem(k+1,2),k=0..n): A005043 := n -> `if`(n=0,1,A001006(n-1)-A005043(n-1)): A189912 := n -> add(binomial(n,k)*A057977(k),k=0..n): A194588 := n -> `if`(n=0,1,A189912(n-1)-A194588(n-1)): A194589 := n -> A194588(n)-A005043(n): # Second method, more efficient: A100071 := n -> A056040(n)*(n/2)^(n-1 mod 2): A194589 := proc(n) local k; (n mod 2)+(1/2)*add((-1)^k*binomial(n,k)*A100071(k+1),k=1..n) end: # Alternatively: a := n -> `if`(n<3,iquo(n,2),hypergeom([1-n/2,-n,3/2-n/2],[1,2-n],4)): seq(simplify(a(n)), n=0..30); # Peter Luschny, Mar 07 2017
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Mathematica
sf[n_] := With[{f = Floor[n/2]}, Pochhammer[f+1, n-f]/f!]; a[n_] := Mod[n, 2] + (1/2)*Sum[(-1)^k*Binomial[n, k]*2^-Mod[k, 2]*(k+1)^Mod[k, 2]*sf[k+1], {k, 1, n}]; Table[a[n], {n, 0, 10}] (* Jean-François Alcover, Jul 30 2013, from 2nd method *) Table[If[n < 3, Quotient[n, 2], HypergeometricPFQ[{1 - n/2, -n, 3/2 - n/2}, {1, 2-n}, 4]], {n,0,30}] (* Peter Luschny, Mar 07 2017 *)
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Maxima
a(n):=sum(binomial(n+2,k)*binomial(n-k,k),k,0,(n)/2); /* Vladimir Kruchinin, Sep 28 2015 */
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PARI
a(n) = sum(k=0, n/2, binomial(n+2,k)*binomial(n-k,k)); vector(30, n, a(n-3)) \\ Altug Alkan, Sep 28 2015
Formula
a(n) = sum_{k=0..n} C(n,k)*A194590(k).
a(n) = (n mod 2)+(1/2)*sum_{k=1..n} (-1)^k*C(n,k)*(k+1)$*((k+1)/2)^(k mod 2). Here n$ denotes the swinging factorial A056040(n).
a(n) = PSUMSIGN([0,0,1,2,6,16,45,..] = PSUMSIGN([0,0,A005717]) where PSUMSIGN is from Sloane's "Transformations of integer sequences". - Peter Luschny, Jan 17 2012
A(x) = B'(x)*(1/x^2-1/(B(x)*x)), where B(x)/x is g.f. of A005043. - Vladimir Kruchinin, Sep 28 2015
a(n) = Sum_{k=0..n/2} C(n+2,k)*C(n-k,k). - Vladimir Kruchinin, Sep 28 2015
a(n) = hypergeom([1-n/2,-n,3/2-n/2],[1,2-n],4) for n>=3. - Peter Luschny, Mar 07 2017
a(n) ~ 3^(n + 1/2) / (8*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 17 2024
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