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.

A095872 Square of the lower triangular matrix T[i,j] = 3j-2 for 1<=j<=i, read by rows.

Original entry on oeis.org

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

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Author

Gary W. Adamson, Jun 10 2004

Keywords

Comments

Arranged by flush left columns (k=1,2,3...), (k=1) column = A000326, the pentagonal numbers (1, 5, 12, 22, 35...). The Octagonal pyramidal number triangle of A095871 is generated from A095872 by dividing the k-th row by the n-th term in the series 1, 4, 7, 10...(k starting with 1). Dividing the 3rd column of A095872 (49, 119, 210, 322, 455...) by 7 generates A059845: 7, 17, 30, 46, 65... Rightmost terms of each row of A095872 are A016778 (1, 16, 49, 100, 169...); i.e. squares of 1, 4, 7, 10... Row sums of A095872 are 1, 21, 105, 325, 780, 1596, 2926... Row sums of A095871 are the octagonal pyramidal numbers, A002414: 1, 9, 30, 70, 135, 231, 364...
[Editor's note: OEIS' "TABL" format (fmt=2) rather displays the transposed matrix as upper triangular matrix.]

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...
		

Crossrefs

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