A125275 Eigensequence of triangle A039599: a(n) = Sum_{k=0..n-1} A039599(n-1,k)*a(k) for n > 0 with a(0) = 1.
1, 1, 2, 7, 31, 162, 968, 6481, 47893, 386098, 3364562, 31460324, 313743665, 3320211313, 37124987124, 436985496790, 5397178181290, 69748452377058, 940762812167126, 13213888481979449, 192891251215160017
Offset: 0
Keywords
Examples
a(3) = 2*(1) + 3*(1) + 1*(2) = 7; a(4) = 5*(1) + 9*(1) + 5*(2) + 1*(7) = 31; a(5) = 14*(1) + 28*(1) + 20*(2) + 7*(7) + 1*(31) = 162. Triangle A039599(n,k) = C(2*n+1, n-k)*(2*k+1)/(2*n+1) (with rows n >= 0 and columns k = 0..n) begins: 1; 1, 1; 2, 3, 1; 5, 9, 5, 1; 14, 28, 20, 7, 1; 42, 90, 75, 35, 9, 1; ... where the g.f. of column k is G000108(x)^(2*k+1) and G000108(x) = (1 - sqrt(1 - 4*x))/(2*x) is the Catalan g.f. function.
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..500
- Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
- Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Programs
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Mathematica
A125275=ConstantArray[0,20]; A125275[[1]]=1; Do[A125275[[n]]=Binomial[2*n-1,n-1]/(2*n-1)+Sum[A125275[[k]]*Binomial[2*n-1,n-k-1]*(2*k+1)/(2*n-1),{k,1,n-1}];,{n,2,20}]; Flatten[{1,A125275}] (* Vaclav Kotesovec, Dec 09 2013 *)
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PARI
a(n)=if(n==0,1,sum(k=0,n-1, a(k)*binomial(2*n-1, n-k-1)*(2*k+1)/(2*n-1)))
Formula
a(n) = Sum_{k=0..n-1} a(k) * C(2*n-1, n-k-1) * (2*k + 1)/(2*n - 1) for n > 0 with a(0) = 1.
Comments