A114040 - OEIS

%I #18 Dec 24 2023 09:37:09

%S 1,9,49,281,1633,9513,55441,323129,1883329,10976841,63977713,

%T 372889433,2173358881,12667263849,73830224209,430314081401,

%U 2508054264193,14618011503753,85200014758321,496582077046169,2894292447518689,16869172608065961,98320743200877073

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

%C 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.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).

%F G.f.: (1+2x-7x^2)/((1-x)(1-6x+x^2)). [_R. J. Mathar_, Sep 09 2008]

%t LinearRecurrence[{7,-7,1},{1,9,49},30] (* _Harvey P. Dale_, Aug 18 2018 *)

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

%K nonn

%O 0,2

%A _N. J. A. Sloane_, based on email from Jack Brennen, Feb 01 2006