A095872 Square of the lower triangular matrix T[i,j] = 3j-2 for 1<=j<=i, read by rows.
1, 5, 16, 12, 44, 49, 22, 84, 119, 100, 35, 136, 210, 230, 169, 51, 200, 322, 390, 377, 256, 70, 176, 455, 580, 624, 560, 361, 70, 276, 455, 580, 624, 560, 361, 92, 364, 609, 800, 910, 912, 779, 484, 117, 464, 784, 1050, 1235, 1312, 1254, 1034, 625, 145, 576, 980, 1330, 1599
Offset: 1
Examples
Let M = the infinite lower triangular matrix in the format exemplified by a 3rd order matrix: [1 0 0 / 1 4 0 / 1 4 7]: i.e. for the n-th order matrix, each row has n terms in the series 1, 4, 7, 10... with the rest of the spaces filled in with zeros. Square the matrix and delete the zeros; then read by rows. [1 0 0 / 1 4 0 / 1 4 7]^2 = [1 0 0 / 5 16 0 / 12 44 49]; then delete the zeros and read by rows: 1, 5, 16, 12, 44, 49...
Programs
-
PARI
A095802(n)={ my( r=sqrtint(2*n)+1, T=matrix(r,r,i,j,if(j>=i,3*j-2))^2); concat(vector(#T,i,vecextract(T[,i],2^i-1)))[n] } \\ M. F. Hasler, Apr 18 2009
Formula
a(k(k+1)/2) = (3k-2)^2 (diagonal elements: squares of the initial series), a(k(k-1)/2+1) = A000326(k) (1st column: pentagonal numbers). - M. F. Hasler, Apr 18 2009
Extensions
Edited and extended by M. F. Hasler, Apr 18 2009
Comments