%I #6 Jan 31 2023 13:55:54
%S 1,1,1,1,2,8,1,3,15,51,1,4,24,120,408,1,5,35,225,1215,4365,1,6,48,372,
%T 2628,15084,53856,1,7,63,567,4851,36603,216405,777609,1,8,80,816,8112,
%U 74352,585792,3558672,12810240,1,9,99,1125,12663,135081,1301157,10623231,65821329,237788055
%N Triangle, read by rows, of coefficients in powers of e.g.f. for A100065 such that, for each row n>=0, Sum_{k=0..n} T(n,k)/k! = [exp(n)] (integer floor of e^n).
%e Rows form the initial coefficients of powers of e.g.f. of A100065:
%e G100065^0: [1,__0,0,0,0,0,0,0,0,...],
%e G100065^1: [1,1,__3,-3,-57,369,3861,-76617,-413775,...],
%e G100065^2: [1,2,8,__12,-84,-12,7200,-40716,-1301328,...],
%e G100065^3: [1,3,15,51,__27,-513,4077,33237,-1211895,...],
%e G100065^4: [1,4,24,120,408,__216,-3168,45576,-202176,...],
%e G100065^5: [1,5,35,225,1215,4365,__1485,-27765,440865,...],
%e G100065^6: [1,6,48,372,2628,15084,53856,__10908,-282960,...],
%e G100065^7: [1,7,63,567,4851,36603,216405,777609,__93177,...],
%e G100065^8: [1,8,80,816,8112,74352,585792,3558672,12810240,...],...
%e such that for each row n, Sum_{k=0..n} T(n,k)/k! = [exp(n)]:
%e [exp(0)] = 1 = 1
%e [exp(1)] = 1+1 = 2
%e [exp(2)] = 1+2+8/2! = 7
%e [exp(3)] = 1+3+15/2!+51/3! = 20
%e [exp(4)] = 1+4+24/2!+120/3!+408/4! = 54
%e [exp(5)] = 1+5+35/2!+225/3!+1215/4!+4365/5! = 148
%o (PARI) {T(n,k)=if(n==0,1,if(k==0,1,if(k==n,n!*(floor(exp(n))-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. A100065.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Nov 02 2004
%E 3 more terms from _Michel Marcus_, Jan 31 2023