%I #14 Jun 13 2017 22:14:23
%S 1,1,4,1,8,26,1,12,63,139,1,16,116,436,726,1,20,185,965,2830,3774,1,
%T 24,270,1790,7335,17634,19601,1,28,371,2975,15505,52444,106827,101784,
%U 1,32,488,4584,28860,124424,358748,633952,528526,1,36,621,6681,49176,256194
%N Triangle, read by rows, of the coefficients of [x^k] in G100234(x)^n such that the row sums are 6^n1 for n>0, where G100234(x) is the g.f. of A100234.
%C 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^nb^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+(m1)*x+sqrt(1+2*(m2*b1)*x+(m^22*m+4*b+1)*x^2))/2; the triangle formed from powers of F(x) will have the g.f.: G(x,y)=(12*x*y+m*x^2*y^2)/((1x*y)*(1(m1)*x*yx^2*y^2x*(1x*y))).
%F G.f.: A(x, y)=(12*x*y+6*x^2*y^2)/((1x*y)*(15*x*yx^2*y^2x*(1x*y))).
%e Rows begin:
%e [1],
%e [1,4],
%e [1,8,26],
%e [1,12,63,139],
%e [1,16,116,436,726],
%e [1,20,185,965,2830,3774],
%e [1,24,270,1790,7335,17634,19601],
%e [1,28,371,2975,15505,52444,106827,101784],
%e [1,32,488,4584,28860,124424,358748,633952,528526],...
%e where row sums form 6^n1 for n>0:
%e 6^11 = 1+4 = 5
%e 6^21 = 1+8+26 = 35
%e 6^31 = 1+12+63+139 = 215
%e 6^41 = 1+16+116+436+726 = 1295
%e 6^51 = 1+20+185+965+2830+3774 = 7775.
%e The main diagonal forms A100236 = [1,4,26,139,726,3774,...],
%e where Sum_{n>=1} A100236(n)/n*x^n = log((1x)/(15*xx^2)).
%t 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}]](* _JeanFrançois Alcover_, May 11 2012, after PARI *)
%o (PARI) T(n,k,m=6)=if(n<k  k<0,0,if(k==0,1, polcoeff(((1+(m1)*x+sqrt(1+2*(m3)*x+(m^22*m+5)*x^2+x*O(x^k)))/2)^n,k)))
%Y Cf. A100234, A100236, A100237, A100232.
%K nonn,tabl
%O 0,3
%A _Paul D. Hanna_, Nov 29 2004
