A125651 Numbers k such that A125650(k) is a perfect square.

1, 3, 24, 147, 864, 5043, 29400, 171363, 998784, 5821347, 33929304, 197754483, 1152597600, 6717831123, 39154389144, 228208503747, 1330096633344, 7752371296323, 45184131144600, 263352415571283, 1534930362283104, 8946229758127347, 52142448186480984

Offset: 1

Author

Alexander Adamchuk, Nov 29 2006

Keywords

• nonn
• easy
• changed

Comments

Corresponding numbers m such that m^2 = A125650(a(n)) are listed in A125652.

3 divides a(n) for n>1. For n>1 a(n) = 3*A001108(n-1), where A001108(k) = {0, 1, 8, 49, 288, 1681, ...}, A001108(k)-th triangular number is a square. - Alexander Adamchuk, Jan 19 2007

Disregarding the term 1, numbers k such that A071910(k) is a nonzero square; i.e., numbers k such that A000096(k) = k*(k+3)/2 is a nonzero square. - Rick L. Shepherd, Jul 13 2012

Examples

			a(2)=3 because A125650(3)=9=3^2; a(3)=24 because A125650(24)=81=9^2.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Crossrefs

Cf. A125650, A125652.

Cf. A001108, A071910, A000096, A000217.

Programs

Magma

I:=[1, 3, 24, 147]; [n le 4 select I[n] else 7*Self(n-1)-7*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, May 21 2012

Mathematica

Join[{1},LinearRecurrence[{7,-7,1},{3,24,147},35]] (* or *) CoefficientList[Series[(-1+x(4+(-10+x)x))/((-1+x)(1+(-6+x) x)),{x,0,35}],x] (* Harvey P. Dale, May 15 2011 *)

Formula

For n>1, a(n+2) = 6*a(n+1) - a(n) + 6.

For n>1, a(n) = ((3+2*sqrt(2))^(n-1) + (3-2*sqrt(2))^(n-1))*3/4 - 3/2.

For n>0, a(2*n) = 3*A002315(n-1)^2; a(2*n+1) = 6*A001542(n)^2.

a(n) = 3*A001108(n-1) for n>1. - Alexander Adamchuk, Jan 19 2007

From Harvey P. Dale, May 15 2011: (Start)

For n>1, a(2)=3, a(3)=24, a(4)=147, a(n) = 7*a(n-1)-7*a(n-2)+a(n-3).

G.f.: x*(-1+x*(4+(-10+x)*x))/((-1+x)*(1+(-6+x)*x)). (End)

Extensions

Edited by Max Alekseyev, Jan 11 2007