

A135880


Triangle P, read by rows, where column k of P^2 equals column 0 of P^(2k+2) such that column 0 of P^2 equals column 0 of P shift one place left, with P(0,0)=1.


23



1, 1, 1, 2, 2, 1, 6, 7, 3, 1, 25, 34, 15, 4, 1, 138, 215, 99, 26, 5, 1, 970, 1698, 814, 216, 40, 6, 1, 8390, 16220, 8057, 2171, 400, 57, 7, 1, 86796, 182714, 93627, 25628, 4740, 666, 77, 8, 1, 1049546, 2378780, 1252752, 348050, 64805, 9080, 1029, 100, 9, 1, 14563135
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OFFSET

0,4


COMMENTS

Amazingly, column 0 (A135881) also equals column 0 of tables A135878 and A135879, both of which have unusual recurrences seemingly unrelated to this triangle.


LINKS



FORMULA

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 Q = A135885 may be defined by
[Q]_k = [P^(2k+2)]_0, for k>=0, such that
(1) Q = P^2 and (2) [Q]_0 = [P]_0 shifted left.
Define the dual triangular matrix R = A135894 by
[R]_k = [P^(2k+1)]_0, for k>=0.
Then columns of P may be formed from powers of R:
[P]_k = [R^(k+1)]_0, for k>=0.
Further, columns of powers of P, Q and R satisfy:
[R^(j+1)]_k = [P^(2k+1)]_j,
[Q^(j+1)]_k = [P^(2k+2)]_j,
[Q^(j+1)]_k = [Q^(k+1)]_j,
[P^(2j+2)]_k = [P^(2k+2)]_j, for all j>=0, k>=0.
Also, we have the column transformations:
R * [P]_k = [P]_{k+1},
Q * [Q]_k = [Q]_{k+1},
Q * [R]_k = [R]_{k+1},
P^2 * [Q]_k = [Q]_{k+1},
P^2 * [R]_k = [R]_{k+1}, for all k>=0.
Other identities include the matrix products:
P^1*R (A135898) = P shifted right one column;
P*R^1*P (A135899) = Q shifted down one row;
R^1*Q (A135900) = R shifted down one row.


EXAMPLE

Triangle P begins:
1;
1, 1;
2, 2, 1;
6, 7, 3, 1;
25, 34, 15, 4, 1;
138, 215, 99, 26, 5, 1;
970, 1698, 814, 216, 40, 6, 1;
8390, 16220, 8057, 2171, 400, 57, 7, 1;
86796, 182714, 93627, 25628, 4740, 666, 77, 8, 1;
1049546, 2378780, 1252752, 348050, 64805, 9080, 1029, 100, 9,
1;
14563135, 35219202, 19003467, 5352788, 1004176, 140908, 15855,
1504, 126, 10, 1;
where column k of P equals column 0 of R^(k+1) where R =
1;
2, 1;
6, 4, 1;
25, 20, 6, 1;
138, 126, 42, 8, 1;
970, 980, 351, 72, 10, 1;
8390, 9186, 3470, 748, 110, 12, 1;
86796, 101492, 39968, 8936, 1365, 156, 14, 1;
1049546, 1296934, 528306, 121532, 19090, 2250, 210, 16, 1; ...
where column k of Q equals column 0 of Q^(k+1) for k>=0;
thus column k of P^2 equals column 0 of P^(2k+2).
1;
1, 1;
2, 3, 1;
6, 12, 5, 1;
25, 63, 30, 7, 1;
138, 421, 220, 56, 9, 1;
970, 3472, 1945, 525, 90, 11, 1;
8390, 34380, 20340, 5733, 1026, 132, 13, 1;
86796, 399463, 247066, 72030, 13305, 1771, 182, 15, 1; ...
where column k of R equals column 0 of P^(2k+1) for k>=0.
Surprisingly, column 0 is also found in triangle A135879:
1;
1,1;
2,2,1,1;
6,6,4,4,2,2,1;
25,25,19,19,13,13,9,5,5,3,1,1;
138,138,113,113,88,88,69,50,50,37,24,24,15,10,5,5,2,1; ...
and is generated by a process that seems completely unrelated.


PROG

(PARI) {T(n, k)=local(P=Mat(1), R, PShR); if(n>0, for(n=0, n, PShR=matrix(#P, #P, r, c, if(r>=c, if(r==c, 1, if(c==1, 0, P[r1, c1])))); R=P*PShR; R=matrix(#P+1, #P+1, r, c, if(r>=c, if(r<#P+1, R[r, c], if(c==1, (P^2)[ #P, 1], (P^(2*c1))[rc+1, 1])))); P=matrix(#R, #R, r, c, if(r>=c, if(r<#R, P[r, c], (R^c)[rc+1, 1]))))); P[n+1, k+1]}


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



