A114040 a(0) = 1, a(1) = 9, a(n) = 6*a(n-1) - a(n-2) - 4.

1, 9, 49, 281, 1633, 9513, 55441, 323129, 1883329, 10976841, 63977713, 372889433, 2173358881, 12667263849, 73830224209, 430314081401, 2508054264193, 14618011503753, 85200014758321, 496582077046169, 2894292447518689, 16869172608065961, 98320743200877073

Offset: 0

Author

N. J. A. Sloane, based on email from Jack Brennen, Feb 01 2006

Keywords

• nonn

Comments

The most straightforward test for "triangularity" is istriangle(n) <===> issquare(8*n+1). If this sequence could be proved to be free of squares beyond the first three terms, that would lead directly to a proof that 0, 1 and 6 are the only triangular numbers whose squares are triangular numbers.

Links

• Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Crossrefs

Equals 8*A001109(n)+1. It is also A081554(n)+1.

Programs

Mathematica

LinearRecurrence[{7,-7,1},{1,9,49},30] (* Harvey P. Dale, Aug 18 2018 *)

Formula

G.f.: (1+2x-7x^2)/((1-x)(1-6x+x^2)). [R. J. Mathar, Sep 09 2008]