A106336 Number of ways of writing n as the sum of n+1 triangular numbers, divided by n+1.
1, 1, 1, 2, 5, 11, 25, 64, 169, 442, 1172, 3180, 8730, 24116, 67159, 188568, 532741, 1512695, 4315996, 12369324, 35587923, 102747636, 297601382, 864525312, 2518185362, 7353088206, 21520084301, 63115752910, 185474840912, 546042990300, 1610314638958
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 5*x^4 + 11*x^5 + 25*x^6 + 64*x^7 +... A(x) = F(x*A(x)) where F(x) = 1 + x + x^3 + x^6 + x^10 + x^15 + x^21 + ... The radius of convergence equals r = 0.322627632692191133... (A106335) at which the g.f. converges to A(r) = 1.987369721184684145... (A106334).
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..2000
Programs
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Maple
b:= proc(n) option remember; expand(`if`(n=0, 1, add(`if`(issqr(8*j+1), x*b(n-j), 0), j=1..n))) end: a:= n-> (p-> add(coeff(p, x, i)*binomial(1+n, i), i=0..n)/(n+1))(b(n)): seq(a(n), n=0..35); # Alois P. Heinz, Jul 31 2017
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Mathematica
f[x_, y_, d_] := f[x, y, d] = If[x < 0 || y < x, 0, If[x == 0 && y == 0, 1, f[x-1, y, 0] + f[x, y - If[d == 0, 1, Ceiling[Sqrt[2*d]]],If[d == 0, 1, Ceiling[Sqrt[2*d]] + d]]]]; Table[f[n, n, 0], {n, 0, 30}] (* David Scambler, May 09 2012 *)
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PARI
{a(n) = my(X); if(n<0,0,X=x+x*O(x^n); polcoef(eta(X^2)^(2*n+2)/eta(X)^(n+1)/(n+1),n))}
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PARI
{a(n) = if(n<0,0,polcoef( sum(k=1,(sqrtint(8*n+1)+1)\2,x^((k^2-k)/2),x*O(x^n))^(n+1)/(n+1),n))}
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PARI
{a(n) = my(A=1+x+x*O(x^n)); for(i=1,n, A=prod(m=1,n,(1+(x*A)^m)*(1-(x*A)^(2*m))));polcoef(A,n)} \\ Paul D. Hanna, Oct 23 2010
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PARI
{a(n) = my(A=1+x); for(i=1,n, A=exp(sum(m=1,n,(x*A)^m/(1+(x*A)^m+x*O(x^n))/m)));polcoef(A,n)} \\ Paul D. Hanna, Jun 01 2011
Formula
G.f.: A(x) = (1/x) * Series_Reversion( x*eta(x)/eta(x^2)^2 ).
G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) A(x) = F(x*A(x)) where F(x) = Sum_{n>=0} x^(n*(n+1)/2).
(2) log(A(x)) = Sum_{n>=1} A106337(n)/n*x^n.
(3) A(x) = Product_{n>=1} (1 + (x*A(x))^n)*(1 - (x*A(x))^(2*n)). - Paul D. Hanna, Oct 23 2010
(4) A(x) = exp( Sum_{n>=1} (x^n*A(x)^n/(1 + x^n*A(x)^n))/n ). - Paul D. Hanna, Jun 01 2011
From Paul D. Hanna, Jun 11 2025: (Start)
(5) A(x)^4 = Sum_{n>=0} (2*n+1) * (x*A(x))^n / (1 - (x*A(x))^(2*n+1)).
(6) A(x^2)^2 = Sum_{n>=0} (x*A(x^2)^(1/2))^n / (1 + (x*A(x^2)^(1/2))^(2*n+1)).
(End)
a(n) ~ c / (n^(3/2) * A106335^n), where c = A366174 = 0.49833479793360342260635926402850016443069428233051290201996853498... - Vaclav Kotesovec, Oct 07 2020
Extensions
Edited by Paul D. Hanna, Jun 01 2011
Comments