A078718 a(n) = (-1)^n*(2*n - 1)*CatalanNumber(n - 2) for n >= 2, a(n) = n for n = 0, 1.
0, 1, 3, -5, 14, -45, 154, -546, 1980, -7293, 27170, -102102, 386308, -1469650, 5616324, -21544100, 82907640, -319929885, 1237518450, -4796857230, 18627909300, -72457790790, 282257178060, -1100982015900, 4299680491080, -16809921068850, 65785111513524, -257683159276956
Offset: 0
Keywords
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
- Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]
Programs
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Mathematica
s = (Sqrt[4*x+1]*(4*x^2+x) + 6*x^2 + x)/(Sqrt[4*x+1] + 4*x+1) + O[x]^28; CoefficientList[s, x] (* Jean-François Alcover, Dec 17 2016, after Vladimir Kruchinin *) Table[(1/2)*Sum[ Binomial[2*n, k]*Binomial[2*(n - 2), 2*n - 2 - k]*(-1)^(2*n - 2 - k), {k, 0, 2*n}], {n, 0, 50}] (* G. C. Greubel, Feb 16 2017 *) (* Mathematica returns CatalanNumber[-2] = 0 and CatalanNumber[-1] = -1. This extended definition is in accordance with the alternative definition of the Catalan numbers C(n) = binomial(2*n, n) - binomial(2*n, n-1). *) a[n_] := (-1)^n (2 n - 1) CatalanNumber[n - 2]; Table[a[n], {n, 0, 21}] (* Peter Luschny, Nov 28 2021 *)
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Maxima
A(x) := x*(1 + sqrt(1+4*x))/2; G(x) := (2*cosh(asinh((3^(3/2)*sqrt(x))/2)/3)*sinh(asinh((3^(3/2)*sqrt(x))/2)/3)* sqrt(x))/(sqrt(3)*sqrt((27*x)/4+1))+(4*sinh(asinh((3^(3/2)*sqrt(x))/2)/3)^2)/3 + 1; taylor(x^2*diff(A(x),x)*G(A(x))/A(x),x,0,20); /* Vladimir Kruchinin, Dec 16 2016 */
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PARI
concat([0,1], for(n=2,25, print1(sum(k=0,2*n, (1/2)* binomial(2*n,k)* binomial(2*( n-2),2*n-k-2)*(-1)^(2*n-k-2)), ", "))) \\ G. C. Greubel, Feb 16 2017
Formula
G.f.: x^2*A'(x)*G(A(x))/A(x), where A(x) = x*(1+sqrt(1+4*x))/2, G(x) =(2*cosh(asinh((3^(3/2)*sqrt(x))/2)/3)*sinh(asinh((3^(3/2)*sqrt(x))/2)/3)*sqrt(x))/(sqrt(3)*sqrt((27*x)/4+1))+(4*sinh(asinh((3^(3/2)*sqrt(x))/2)/3)^2)/3+1. - Vladimir Kruchinin, Dec 16 2016
G.f.: (sqrt(4*x+1)*(4*x^2+x)+6*x^2+x)/(sqrt(4*x+1)+4*x+1). - Vladimir Kruchinin, Dec 17 2016
a(n) ~ (-1)^n * 2^(2*n-3) / sqrt(Pi*n). - Vaclav Kotesovec, Dec 17 2016
a(n) = (1/2)*Sum_{k=0,..,2*n} ( binomial(2*n, k)*binomial(2*(n - 2), 2*n - 2 - k)*(-1)^(2*n - 2 - k) ), with a(0)=0, a(1)=1. - G. C. Greubel, Feb 16 2017
D-finite with recurrence (-n+1)*a(n) +6*(-2*n+5)*a(n-1) +16*(-2*n+7)*a(n-2)=0. - R. J. Mathar, Nov 22 2024
Extensions
New name by Peter Luschny, Nov 28 2021
Comments