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FORMULA
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Denote this triangle by P and define as follows.
Let [P^m]_k denote column k of matrix power P^m,
so that triangular matrix W = A136231 may be defined by
[W]_k = [P^(3k+3)]_0, for k>=0, such that
(1) W = P^3 and (2) [W]_0 = [P]_0 shift up one row.
Define the triangular matrix U = A136228 by
[U]_k = [P^(3k+1)]_0, for k>=0,
and the triangular matrix V = A136230 by
[V]_k = [P^(3k+2)]_0, for k>=0.
Then columns of P may be formed from powers of U:
[P]_k = [U^(k+1)]_0, for k>=0,
and columns of P^2 may be formed from powers of V:
[P^2]_k = [V^(k+1)]_0, for k>=0.
Further, columns of powers of P, U, V and W satisfy:
[U^(j+1)]_k = [P^(3k+1)]_j,
[V^(j+1)]_k = [P^(3k+2)]_j,
[W^(j+1)]_k = [P^(3k+3)]_j,
[W^(j+1)]_k = [W^(k+1)]_j,
[P^(3j+3)]_k = [P^(3k+3)]_j, for all j>=0, k>=0.
Also, we have the column transformations:
U * [P]_k = [P]_{k+1},
V * [P^2]_k = [P^2]_{k+1},
W * [P^3]_k = [P^3]_{k+1},
W * [U]_k = [U]_{k+1},
W * [V]_k = [V]_{k+1},
W * [W]_k = [W]_{k+1}, for all k>=0.
Other identities include the matrix products:
U = P * [P^2 shift right one column];
V = P^2 * [P shift right one column];
V = U * [U shift down one row];
W = V * [V shift down one row];
where the triangle transformations "shift right" and "shift down" are illustrated in examples of entries A136228 (U) and A136230 (V).
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EXAMPLE
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Triangle P begins:
1;
1, 1;
3, 2, 1;
15, 10, 3, 1;
108, 75, 21, 4, 1;
1036, 753, 208, 36, 5, 1;
12569, 9534, 2637, 442, 55, 6, 1;
185704, 146353, 40731, 6742, 805, 78, 7, 1;
3247546, 2647628, 742620, 122350, 14330, 1325, 105, 8, 1;
65762269, 55251994, 15624420, 2571620, 298240, 26943, 2030, 136, 9, 1; ...
where column k of P = column 0 of U^(k+1) and U = A136228.
Matrix cube, W = P^3 (A136231), begins:
1;
3, 1;
15, 6, 1;
108, 48, 9, 1;
1036, 495, 99, 12, 1;
12569, 6338, 1323, 168, 15, 1;
185704, 97681, 21036, 2754, 255, 18, 1; ...
where column k of P^3 = column 0 of P^(3k+3) such that
column 0 of P^3 = column 0 of P shift one row up.
Matrix square, P^2 (A136225), begins:
1;
2, 1;
8, 4, 1;
49, 26, 6, 1;
414, 232, 54, 8, 1;
4529, 2657, 629, 92, 10, 1;
61369, 37405, 9003, 1320, 140, 12, 1; ...
where column k of P^2 = column 0 of V^(k+1) and
1;
2, 1;
8, 5, 1;
49, 35, 8, 1;
414, 325, 80, 11, 1;
4529, 3820, 988, 143, 14, 1;
61369, 54800, 14696, 2200, 224, 17, 1; ...
where column k of V = column 0 of P^(3k+2).
Related triangle U = A136228 begins:
1;
1, 1;
3, 4, 1;
15, 24, 7, 1;
108, 198, 63, 10, 1;
1036, 2116, 714, 120, 13, 1;
12569, 28052, 9884, 1725, 195, 16, 1; ...
where column k of U = column 0 of P^(3k+1)
and column k of P = column 0 of U^(k+1).
Surprisingly, column 0 of P is also found in square A136217:
(1),(1),1,(1),1,(1),1,(1),1,1,(1),1,1,(1),1,1,(1),1,1,1,(1),...;
(1),(2),3,(4),5,(6),7,(8),9,10,(11),12,13,(14),15,16,(17),...;
(3),(8),15,(24),34,(46),59,(74),90,108,(127),147,169,(192),...;
(15),(49),108,(198),306,(453),622,(838),1080,1377,(1704),...;
(108),(414),1036,(2116),3493,(5555),8040,(11477),15483,...;
(1036),(4529),12569,(28052),48800,(82328),124335,(186261),...;
(12569),(61369),185704,(446560),811111,(1438447),2250731,...;
...
and has a recurrence similar to that of square array A136212
which generates the triple factorials.
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PROG
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(PARI) {T(n, k)=local(P=Mat(1), U, PShR); if(n>0, for(i=0, n, PShR=matrix(#P, #P, r, c,
if(r>=c, if(r==c, 1, if(c==1, 0, P[r-1, c-1])))); U=P*PShR^2; U=matrix(#P+1,
#P+1, r, c, if(r>=c, if(r<#P+1, U[r, c], if(c==1, (P^3)[ #P, 1], (P^(3*c-1))[r-c+1,
1])))); P=matrix(#U, #U, r, c, if(r>=c, if(r<#R, P[r, c], (U^c)[r-c+1,
1]))))); P[n+1, k+1]}
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