%I
%S 1,2,1,3,4,2,4,11,14,6,5,26,64,66,24,6,57,244,456,384,120,7,120,846,
%T 2556,3744,2640,720,8,247,2778,12762,28944,34560,20880,5040,9,502,
%U 8828,59382,195768,352080,353520,186480,40320,10,1013,27488,264012,1216368,3091320,4587120,3966480,1854720,362880
%N Triangle read by rows, related to A055129 (repunits in base k).
%C Second column of A107893 = Eulerian numbers (A000295) starting with 1: 1, 4, 11, 26, 57, ... Rightmost term in row n = (n1)!.
%C Using the Jun 18 2009 formula of _Johannes W. Meijer_ in A028246: Instead of a(n,1)=1 set a(n,1)=n. The result is A107893.  _Werner Schulte_, Dec 12 2016
%H Michael De Vlieger, <a href="/A107893/b107893.txt">Table of n, a(n) for n = 1..5050</a> (1 <= n <= 100)
%F nth row = inverse binomial transform of nth column of A055129, where the latter are generated from f(x) = x^(n1) + x^(n2) + ...+ x + 1; (x = 1, 2, 3, ...)
%F A(n,k) = Sum_{i=1..n} A028246(i,k) for 1 <= k <= n.  _Werner Schulte_, Dec 08 2016
%F The polynomials p(n,t) = Sum_{k=1..n} A(n,k)*t^k are given by p(1,t) = t and p(n+1,t) = t + t*(t+1)*(d/dt)p(n,t) for n >= 1.  _Werner Schulte_, Dec 12 2016
%e Binomial transform of Row 4 in the form: (4, 11, 14, 6, 0, 0, 0, ...) = Row 4 of A055129: 4, 15, 40, 85, ... which is generated from f(x) = x^3 + x^2 + x + 1; (x = 1,2,3, ...).
%e Triangle starts:
%e 1;
%e 2, 1;
%e 3, 4, 2;
%e 4, 11, 14, 6;
%e 5, 26, 64, 66, 24;
%e 6, 57, 244, 456, 384, 120;
%e ...
%t Table[Sum[Sum[(1)^(k  j) Binomial[k, j] j^i, {j, 0, k}]/k, {i, n}], {n, 10}, {k, n}] // Flatten (* _Michael De Vlieger_, Dec 11 2016, after _JeanFrançois Alcover_ at A028246 *)
%Y Cf. A000295, A055129, A028246.
%K nonn,tabl
%O 1,2
%A _Gary W. Adamson_, May 26 2005
