cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A136225 Matrix square of triangle A136220, read by rows.

Original entry on oeis.org

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, 996815, 627435, 153276, 22606, 2385, 198, 14, 1, 18931547, 12248365, 3031553, 450066, 47500, 3904, 266, 16, 1, 412345688, 273211787
Offset: 0

Views

Author

Paul D. Hanna, Jan 28 2008

Keywords

Comments

Column 0 of this triangle = column 1 of square array A136217.

Examples

			Let P = A136220, then this triangle is P^2 and 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;
996815, 627435, 153276, 22606, 2385, 198, 14, 1;
18931547, 12248365, 3031553, 450066, 47500, 3904, 266, 16, 1; ...
where column k of P^2 = column 0 of V^(k+1) and
triangle V = A136230 begins:
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).
Triangle P = A136220 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; ...
where column k of P = column 0 of U^(k+1) and U = A136228.
		

Crossrefs

Cf. columns: A136226, A136227; related tables: A136228 (U), A136230 (V), A136231 (W=P^3), A136217, A136218.

Programs

  • 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^2)[n+1,k+1]}

Formula

Let P=A136220, V=A136230, then column k of P^2 (this triangle) = column 0 of V^(k+1) while column j of V = column 0 of P^(3j+2).