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Triangle, read by rows, where the n-th row lists the coefficients of the polynomial of degree n, with root -1, that generates the n-th diagonal of this sequence.
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%I #14 Sep 21 2015 10:24:07

%S 1,1,1,1,2,1,2,4,3,1,4,10,9,4,1,10,28,30,16,5,1,30,90,108,68,25,6,1,

%T 106,328,426,304,130,36,7,1,420,1338,1842,1444,700,222,49,8,1,1818,

%U 6024,8706,7320,3930,1404,350,64,9,1,8530,29626,44736,39700,23110,9150,2548

%N Triangle, read by rows, where the n-th row lists the coefficients of the polynomial of degree n, with root -1, that generates the n-th diagonal of this sequence.

%C The leftmost column (A091174) is determined by the condition that the root of each row polynomial is -1. The next column is T(n,1)=A091175(n+1) (n>=0).

%H Paul D. Hanna, <a href="/A091173/b091173.txt">Table of n, a(n) for n = 0..1034</a>

%F T(n+k, k) = Sum_{j=0..n} T(n, j) * k^j, with T(0,0)=1, T(0,n)=1 and T(n,0) = -Sum_{j=1..n} T(n, j) * (-1)^j.

%e For n=3, k=2, T(n+k,k) = T(5,2) = 30 = (2) + (4)2 + (3)2^2 + (1)2^3.

%e For n=4, k=3, T(n+k,k) = T(7,3) = 304 = (4) + (10)3 + (9)3^2 + (4)3^3 + (1)3^4.

%e Rows begin with n=0:

%e 1;

%e 1, 1;

%e 1, 2, 1;

%e 2, 4, 3, 1;

%e 4, 10, 9, 4, 1;

%e 10, 28, 30, 16, 5, 1;

%e 30, 90, 108, 68, 25, 6, 1;

%e 106, 328, 426, 304, 130, 36, 7, 1;

%e 420, 1338, 1842, 1444, 700, 222, 49, 8, 1;

%e 1818, 6024, 8706, 7320, 3930, 1404, 350, 64, 9, 1;

%e 8530, 29626, 44736, 39700, 23110, 9150, 2548, 520, 81, 10, 1;

%e 43430, 158012, 248466, 230424, 142890, 61680, 18970, 4288, 738, 100, 11, 1;

%e 240208, 909010, 1483398, 1429236, 931500, 431646, 144858, 35976, 6804, 1010, 121, 12, 1; ...

%t T[0, _] = 1; T[n_, 0] := T[n, 0] = -Sum[T[n, j]*(-1)^j, {j, 1, n}]; T[n_, k_] := T[n, k] = Sum[T[n-k, j]*k^j, {j, 0, n-k}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Sep 21 2015 *)

%o (PARI) {T(n,k)=if(n==k,1,if(n>k&k>0,sum(j=0,n-k,T(n-k,j)*k^j),if(k==0,-sum(j=1,n,T(n,j)*(-1)^j))))}

%o for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

%Y Cf. A091174, A091175.

%K nonn,tabl

%O 0,5

%A _Paul D. Hanna_, Dec 25 2003