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A100229 Triangle, read by rows, of the coefficients of [x^k] in G100228(x)^n such that the row sums are 4^n-1 for n>0, where G100228(x) is the g.f. of A100228.

%I A100229 #7 Jun 13 2017 22:14:31
%S A100229 1,1,2,1,4,10,1,6,21,35,1,8,36,92,118,1,10,55,185,380,392,1,12,78,322,
%T A100229 879,1506,1297,1,14,105,511,1715,3948,5803,4286,1,16,136,760,3004,
%U A100229 8536,17020,21904,14158,1,18,171,1077,4878,16344,40395,71109,81387,46763
%N A100229 Triangle, read by rows, of the coefficients of [x^k] in G100228(x)^n such that the row sums are 4^n-1 for n>0, where G100228(x) is the g.f. of A100228.
%C A100229 The main diagonal forms A100230. Secondary diagonal is T(n+1,n) = (n+1)*A052924(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))).
%F A100229 G.f.: A(x, y)=(1-2*x*y+4*x^2*y^2)/((1-x*y)*(1-3*x*y-x^2*y^2-x*(1-x*y))).
%e A100229 Rows begin:
%e A100229 [1],
%e A100229 [1,2],
%e A100229 [1,4,10],
%e A100229 [1,6,21,35],
%e A100229 [1,8,36,92,118],
%e A100229 [1,10,55,185,380,392],
%e A100229 [1,12,78,322,879,1506,1297],
%e A100229 [1,14,105,511,1715,3948,5803,4286],
%e A100229 [1,16,136,760,3004,8536,17020,21904,14158],...
%e A100229 where row sums form 4^n-1 for n>0:
%e A100229 4^1-1 = 1+2 = 3
%e A100229 4^2-1 = 1+4+10 = 15
%e A100229 4^3-1 = 1+6+21+35 = 63
%e A100229 4^4-1 = 1+8+36+92+118 = 255
%e A100229 4^5-1 = 1+10+55+185+380+392 = 1023.
%e A100229 The main diagonal forms A100230 = [1,2,10,35,118,392,1297,...],
%e A100229 where Sum_{n>=1} A100230(n)/n*x^n = log((1-x)/(1-3*x-x^2)).
%o A100229 (PARI) T(n,k,m=4)=if(n<k || k<0,0,if(k==0,1, polcoeff(((1+(m-1)*x+sqrt(1+2*(m-3)*x+(m^2-2*m+5)*x^2+x*O(x^k)))/2)^n,k)))
%Y A100229 Cf. A100228, A100230, A052924.
%K A100229 nonn,tabl
%O A100229 0,3
%A A100229 _Paul D. Hanna_, Nov 29 2004