A107671 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, ...}.

1, 8, 2, 513, 27, 3, 81856, 2368, 64, 4, 23846125, 469625, 7625, 125, 5, 10943504136, 160767720, 1898856, 19656, 216, 6, 7250862593527, 83548607478, 776598305, 6081733, 43561, 343, 7, 6545029128786432, 61068815111168, 465690017280, 2966844928, 16494080, 86528, 512, 8

Offset: 0

Author

Paul D. Hanna, Jun 07 2005

Keywords

• nonn
• tabl

Examples

			Triangle T begins:
              1;
              8,           2;
            513,          27,         3;
          81856,        2368,        64,       4;
       23846125,      469625,      7625,     125,     5;
    10943504136,   160767720,   1898856,   19656,   216,   6;
  7250862593527, 83548607478, 776598305, 6081733, 43561, 343, 7;
  ...
The matrix cube T^3 shifts each row to the right 1 place, dropping the diagonal D and putting A006690 in column 0:
             1;
            56,           8;
          7965,         513,        27;
       2128064,       81856,      2368,      64;
     914929500,    23846125,    469625,    7625,  125;
  576689214816, 10943504136, 160767720, 1898856, 19656, 216;
  ...
From _Petros Hadjicostas_, Mar 11 2021: (Start)
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).
The value of the denominator Product_{j=1..r} s_j! for these four terms is 6, 2, 2, and 1, respectively.
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.
Thus T(2,0) = (19683/3)/6 - (729/1)/2 - (1728/2)/2 + (216/1)/1 = 513. (End)
		

Crossrefs

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

Programs

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])}

Formula

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.

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