

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, ...}.


6



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
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OFFSET

0,2


LINKS

Table of n, a(n) for n=0..35.


FORMULA

Matrix diagonalization method: define the triangular matrix P by P(n, k) = ((n+1)^3)^(nk)/(nk)! 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)^(r1) * 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


EXAMPLE

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)


PROG

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


CROSSREFS

Cf. A107667, A107672 (column 0), A107673, A107674 (matrix square), A107676 (matrix cube), A006690.
Sequence in context: A176860 A281068 A093082 * A271174 A216891 A256783
Adjacent sequences: A107668 A107669 A107670 * A107672 A107673 A107674


KEYWORD

nonn,tabl


AUTHOR

Paul D. Hanna, Jun 07 2005


STATUS

approved



