A100235 Triangle, read by rows, of the coefficients of [x^k] in G100234(x)^n such that the row sums are 6^n-1 for n>0, where G100234(x) is the g.f. of A100234.

1, 1, 4, 1, 8, 26, 1, 12, 63, 139, 1, 16, 116, 436, 726, 1, 20, 185, 965, 2830, 3774, 1, 24, 270, 1790, 7335, 17634, 19601, 1, 28, 371, 2975, 15505, 52444, 106827, 101784, 1, 32, 488, 4584, 28860, 124424, 358748, 633952, 528526, 1, 36, 621, 6681, 49176, 256194

Offset: 0

Author

Paul D. Hanna, Nov 29 2004

Keywords

• nonn
• tabl

Comments

The main diagonal forms A100236. Secondary diagonal is: T(n+1,n) = (n+1)*A100237(n). 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],
[1,4],
[1,8,26],
[1,12,63,139],
[1,16,116,436,726],
[1,20,185,965,2830,3774],
[1,24,270,1790,7335,17634,19601],
[1,28,371,2975,15505,52444,106827,101784],
[1,32,488,4584,28860,124424,358748,633952,528526],...
where row sums form 6^n-1 for n>0:
6^1-1 = 1+4 = 5
6^2-1 = 1+8+26 = 35
6^3-1 = 1+12+63+139 = 215
6^4-1 = 1+16+116+436+726 = 1295
6^5-1 = 1+20+185+965+2830+3774 = 7775.
The main diagonal forms A100236 = [1,4,26,139,726,3774,...],
where Sum_{n>=1} A100236(n)/n*x^n = log((1-x)/(1-5*x-x^2)).
		

Crossrefs

Cf. A100234, A100236, A100237, A100232.

Programs

Mathematica

row[n_] := CoefficientList[ Series[ (1 + 5*x + Sqrt[1 + 6*x + 29*x^2])^n/2^n, {x, 0, n}], x]; Flatten[ Table[ row[n], {n, 0, 9}]](* Jean-François Alcover, May 11 2012, after PARI *)

PARI

T(n,k,m=6)=if(n
				

Formula

G.f.: A(x, y)=(1-2*x*y+6*x^2*y^2)/((1-x*y)*(1-5*x*y-x^2*y^2-x*(1-x*y))).