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Numbers n such that 3*T(n)+1 is a square, where T = A000217.
5

%I #30 Jun 13 2015 00:54:55

%S 0,1,6,15,64,153,638,1519,6320,15041,62566,148895,619344,1473913,

%T 6130878,14590239,60689440,144428481,600763526,1429694575,5946945824,

%U 14152517273,58868694718,140095478159,582740001360,1386802264321,5768531318886,13727927165055

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

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

%H Bruno Berselli, <a href="/A233450/b233450.txt">Table of n, a(n) for n = 1..200</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,10,-10,-1,1).

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

%F 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.

%F 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.

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

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

%Y Sequence A129444 gives n+1.

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

%Y 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.

%K nonn,easy

%O 1,3

%A _Bruno Berselli_, Dec 10 2013