A233450 Numbers n such that 3*T(n)+1 is a square, where T = A000217.

0, 1, 6, 15, 64, 153, 638, 1519, 6320, 15041, 62566, 148895, 619344, 1473913, 6130878, 14590239, 60689440, 144428481, 600763526, 1429694575, 5946945824, 14152517273, 58868694718, 140095478159, 582740001360, 1386802264321, 5768531318886, 13727927165055

Offset: 1

Author

Bruno Berselli, Dec 10 2013

Keywords

• nonn
• easy

Comments

For n>1, partial sums of A080872 starting from A080872(1).

Examples

			153 is in the sequence because 3*153*154/2+1 = 188^2.
		

Links

• Bruno Berselli, Table of n, a(n) for n = 1..200
• Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

Crossrefs

Sequence A129444 gives n+1.

Cf. A000217, A080872, A129445 (square roots of 3*A000217(a(n))+1), A132596 (numbers m such that 3*A000217(m) is a square).

Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; this sequence for k=3; A001652 for k=4; A129556 for k=5; A001921 for k=6.

Programs

Mathematica

LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 6, 15, 64}, 30]

Formula

G.f.: x^2*(1 + 5*x - x^2 - x^3) / ((1 - x)*(1 - 10*x^2 + x^4)).

a(n) = a(n-1) +10*a(n-2) -10*a(n-3) -a(n-4) +a(n-5) for n>5, a(1)=0, a(2)=1, a(3)=6, a(4)=15, a(5)=64.

a(n) = -1/2 + ( (-3*(-1)^n + 2*sqrt(6))*(5 + 2*sqrt(6))^floor(n/2) - (3*(-1)^n + 2*sqrt(6))*(5 - 2*sqrt(6))^floor(n/2) )/12.