A107671 - OEIS

%I #21 Mar 11 2021 07:43:25

%S 1,8,2,513,27,3,81856,2368,64,4,23846125,469625,7625,125,5,

%T 10943504136,160767720,1898856,19656,216,6,7250862593527,83548607478,

%U 776598305,6081733,43561,343,7,6545029128786432,61068815111168,465690017280,2966844928,16494080,86528,512,8

%N Triangular matrix T, read by rows, that satisfies: T = D + SHIFT_LEFT(T^3), where SHIFT_LEFT shifts each row 1 place to the left and D is the diagonal matrix {1, 2, 3, ...}.

%F Matrix diagonalization method: define the triangular matrix P by P(n, k) = ((n+1)^3)^(n-k)/(n-k)! for n >= k >= 0 and the diagonal matrix D(n, n) = n+1 for n >= 0; then T is given by T = P^-1*D*P.

%F T(n,k=0) = Sum_{r=1..(n+1)} (-1)^(r-1) * Sum_{s_1, ..., s_r} (s_1^(-1)/(Product_{j=1..r} s_j!)) * Product_{j=1..r} (Sum_{i=1..j} s_i)^(3*s_j)), where the second sum is over lists (s_1, ..., s_r) of positive integers s_i such that Sum_{i=1..r} s_i = n + 1. (Thus, the second sum is over all compositions of n + 1.) - _Petros Hadjicostas_, Mar 11 2021

%e Triangle T begins:

%e               1;

%e               8,           2;

%e             513,          27,         3;

%e           81856,        2368,        64,       4;

%e        23846125,      469625,      7625,     125,     5;

%e     10943504136,   160767720,   1898856,   19656,   216,   6;

%e   7250862593527, 83548607478, 776598305, 6081733, 43561, 343, 7;

%e   ...

%e The matrix cube T^3 shifts each row to the right 1 place, dropping the diagonal D and putting A006690 in column 0:

%e              1;

%e             56,           8;

%e           7965,         513,        27;

%e        2128064,       81856,      2368,      64;

%e      914929500,    23846125,    469625,    7625,  125;

%e   576689214816, 10943504136, 160767720, 1898856, 19656, 216;

%e   ...

%e From _Petros Hadjicostas_, Mar 11 2021: (Start)

%e We illustrate the above formula for T(n,k=0) with the compositions of n + 1 for n = 2. The compositions of n + 1 = 3 are 3, 1 + 2, 2 + 1, and 1 + 1 + 1.  Thus the above sum has four terms with (r = 1, s_1 = 3), (r = 2, s_1 = 1, s_2 = 2), (r = 2, s_1 = 2, s_2 = 1), and (r = 3, s_1 = s_2 = s_3 = 1).

%e The value of the denominator Product_{j=1..r} s_j! for these four terms is 6, 2, 2, and 1, respectively.

%e The value of the numerator s_1^(-1)*Product_{j=1..r} (Sum_{i=1..j} s_i)^(3*s_j) for these four terms is 19683/3, 729/1, 1728/2, and 216/1.

%e Thus T(2,0) = (19683/3)/6 - (729/1)/2 - (1728/2)/2 + (216/1)/1 = 513. (End)

%o (PARI) {T(n,k)=local(P=matrix(n+1,n+1,r,c,if(r>=c,(r^3)^(r-c)/(r-c)!)), D=matrix(n+1,n+1,r,c,if(r==c,r)));if(n>=k,(P^-1*D*P)[n+1,k+1])}

%Y Cf. A107667, A107672 (column 0), A107673, A107674 (matrix square), A107676 (matrix cube), A006690.

%K nonn,tabl

%O 0,2

%A _Paul D. Hanna_, Jun 07 2005