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%I #3 Mar 30 2012 18:36:43
%S 1,1,1,1,2,4,1,3,9,15,1,4,16,48,96,1,5,25,105,345,555,1,6,36,192,864,
%T 2772,6408,1,7,49,315,1785,8169,28665,59157,1,8,64,480,3264,19056,
%U 91296,323424,734976,1,9,81,693,5481,38583,233361,1144611,4222449,9129591,1
%N Triangle, read by rows, of coefficients in powers of e.g.f. for A100076 such that, for each row n>=0, Sum_{k=0..n} T(n,k)/k! = [sqrt(5)^n].
%e Rows form the initial coefficients of powers of e.g.f. of A100076:
%e G100076^0: [1,__0,0,0,0,0,0,0,0,...],
%e G100076^1: [1,1,__1,-3,9,-33,513,-10917,155313,...],
%e G100076^2: [1,2,4,__0,0,-36,1080,-17928,181440,...],
%e G100076^3: [1,3,9,15,__9,-99,1521,-17631,112401,...],
%e G100076^4: [1,4,16,48,96,__-72,1296,-11664,31104,...],
%e G100076^5: [1,5,25,105,345,555,__1305,-6705,-6255,...],
%e G100076^6: [1,6,36,192,864,2772,6408,__648,-15552,...],
%e G100076^7: [1,7,49,315,1785,8169,28665,59157,__41265,...],...
%e such that for each row n, Sum_{k=0..n} T(n,k)/k! = [sqrt(5)^n]:
%e [sqrt(5)^0] = 1 = 1
%e [sqrt(5)^1] = 1+1 = 2
%e [sqrt(5)^2] = 1+2+4/2! = 5
%e [sqrt(5)^3] = 1+3+9/2!+15/3! = 11
%e [sqrt(5)^4] = 1+4+16/2!+48/3!+96/4! = 25
%e [sqrt(5)^5] = 1+5+25/2!+105/3!+345/4!+555/5! = 55
%e [sqrt(5)^6] = 1+6+36/2!+192/3!+864/4!+2772/5!+6408/6! = 125
%e [sqrt(5)^7] = 1+7+49/2!+315/3!+1785/4!+8169/5!+28665/6!+59157/7! = 279
%o (PARI) {T(n,k)=if(n==0,1,if(k==0,1,if(k==n,n!*(floor(sqrt(5)^n+1/10^15)-sum(j=0,n-1,T(n,j)/j!)), k!*polcoeff((Ser(vector(n,i,T(n-1,i-1)/(i-1)!),x)+x*O(x^k))^(n/(n-1)),k,x))))}
%Y Cf. A100076, A100064.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Nov 03 2004
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