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 A078341 Triangle read by rows: T(n,k) = n*T(n-1,k-1) + k*T(n-1,k) starting with T(0,0)=1. 1

%I

%S 1,0,1,0,1,2,0,1,7,6,0,1,18,46,24,0,1,41,228,326,120,0,1,88,930,2672,

%T 2556,720,0,1,183,3406,17198,31484,22212,5040,0,1,374,11682,96040,

%U 295004,385144,212976,40320,0,1,757,38412,489298,2339380,4965900

%N Triangle read by rows: T(n,k) = n*T(n-1,k-1) + k*T(n-1,k) starting with T(0,0)=1.

%C Triangle of coefficients of polynomials P[n]. Let F(t) satisfy dF/dt = exp(x*(exp(F)-1)) and F(0)=0. Then F(t) = Sum_{n>=0} P[n]/n! t^n, where P[n] is a polynomial in x of degree n-1. The constant term of the polynomial is zero for n >= 2. The coefficient of x is 1 for n >= 2. The coefficient of x^n in P[n+1] is n!. The value at 1 is given by sequence A007549.

%F P[1]=1; P[n+1] = x*(d/dx)P[n] + x*n*P[n].

%e P[1]=1, P[2]=x, P[3]=x+2*x^2, P[4]=x+7*x^2+6*x^3, P[5]=x+18*x^2+46*x^3+24*x^4, P[6]=x+41*x^2+228*x^3+326*x^4+120*x^5.

%e Rows start 1; 0,1; 0,1,2; 0,1,7,6; 0,1,18,46,24; 0,1,41,228,326,120; ...

%p P[1] := 1; for n from 1 to 10 do P[n+1] := expand(x*diff(P[n],x)+x*n*P[n]) od;

%t p[1][x_] = 1; p[n_][x_] := x*p[n-1]'[x] + x*(n-1)*p[n-1][x]; Table[ CoefficientList[ p[n][x], x], {n, 1, 10}] // Flatten (* _Jean-François Alcover_, Jan 29 2013 *)

%Y Cf. A007549, A000142.

%Y Columns include A000007, A057427, A095151, A103768. Diagonals include A000142, A067318. Row sums are A007549.

%K easy,nonn,tabl

%O 1,6

%A _F. Chapoton_, Nov 22 2002

%E Additional comments from _Henry Bottomley_, Feb 15 2005

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Last modified August 6 15:06 EDT 2020. Contains 336248 sequences. (Running on oeis4.)