A100232 Triangle, read by rows, of the coefficients of [x^k] in G100231(x)^n such that the row sums are 5^n-1 for n>0, where G100231(x) is the g.f. of A100231.

1, 1, 3, 1, 6, 17, 1, 9, 39, 75, 1, 12, 70, 220, 321, 1, 15, 110, 470, 1165, 1363, 1, 18, 159, 852, 2895, 5922, 5777, 1, 21, 217, 1393, 5943, 16807, 29267, 24475, 1, 24, 284, 2120, 10822, 38536, 93468, 141688, 103681, 1, 27, 360, 3060, 18126, 77274, 236748

Offset: 0

Author

Paul D. Hanna, Nov 29 2004

Keywords

• nonn
• tabl

Comments

The main diagonal forms A100233. Secondary diagonal is: T(n+1,n) = (n+1)*A033887(n) = (n+1)*Fibonacci(3*n+1). More generally, if g.f. F(x) satisfies: m^n-b^n = Sum_{k=0..n} [x^k]F(x)^n, then F(x) also satisfies: (m+z)^n - (b+z)^n + z^n = Sum_{k=0..n} [x^k](F(x)+z*x)^n for all z and F(x)=(1+(m-1)*x+sqrt(1+2*(m-2*b-1)*x+(m^2-2*m+4*b+1)*x^2))/2; the triangle formed from powers of F(x) will have the g.f.: G(x,y)=(1-2*x*y+m*x^2*y^2)/((1-x*y)*(1-(m-1)*x*y-x^2*y^2-x*(1-x*y))).

Examples

			Rows begin:
[1],
[1,3],
[1,6,17],
[1,9,39,75],
[1,12,70,220,321],
[1,15,110,470,1165,1363],
[1,18,159,852,2895,5922,5777],
[1,21,217,1393,5943,16807,29267,24475],
[1,24,284,2120,10822,38536,93468,141688,103681],...
where row sums form 5^n-1 for n>0:
5^1-1 = 1+3 = 4
5^2-1 = 1+6+17 = 24
5^3-1 = 1+9+39+75 = 124
5^4-1 = 1+12+70+220+321 = 624
5^5-1 = 1+15+110+470+1165+1363 = 3124.
The main diagonal forms A100233 = [1,3,17,75,321,1363,5777,...],
where Sum_{n>=1} A100233(n)/n*x^n = log((1-x)/(1-4*x-x^2)).
		

Crossrefs

Cf. A100231, A100233, A033887, A100229.

Programs

PARI

T(n,k,m=5)=if(n
				

Formula

G.f.: A(x, y)=(1-2*x*y+5*x^2*y^2)/((1-x*y)*(1-4*x*y-x^2*y^2-x*(1-x*y))).