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Numbers m such that 5*T(m)-1 is a square, where T = A000217.
2

%I #22 Nov 30 2016 05:34:39

%S 1,4,52,169,1993,6436,75700,244417,2874625,9281428,109160068,

%T 352449865,4145207977,13383813460,157408743076,508232461633,

%U 5977387028929,19299449728612,226983298356244,732870857225641,8619387950508361,27829793124845764

%N Numbers m such that 5*T(m)-1 is a square, where T = A000217.

%C Sum_{i=1..infinity} 1/a(i) = 1.2758228304947598524736181699610...

%H Bruno Berselli, <a href="/A233474/b233474.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 (1,38,-38,-1,1)

%F G.f.: x*(1 + 3*x + 10*x^2 + 3*x^3 + x^4) / ((1 - x)*(1 - 38*x^2 + x^4)).

%F a(n) = a(n-1) + 38*a(n-2) - 38*a(n-3) - a(n-4) + a(n-5) for n>5, a(1)=1, a(2)=4, a(3)=52, a(4)=169, a(5)=1993.

%F a(n) = -1/2 + ( (15 + 4*sqrt(10)*(-1)^n)*(19 - 6*sqrt(10))^floor(n/2) + (15 - 4*sqrt(10)*(-1)^n)*(19 + 6*sqrt(10))^floor(n/2) )/20.

%e 169 is in the sequence because 5*169*170/2-1 = 268^2.

%t LinearRecurrence[{1, 38, -38, -1, 1}, {1, 4, 52, 169, 1993}, 30]

%Y Cf. A000217, A129556 (numbers m such that 5*A000217(m)+1 is a square), A132593 (numbers m such that 5*A000217(m) is a square).

%Y Cf. numbers m such that k*A000217(m)-1 is a square: A072221 for k=1; m=1 for k=2; this sequence for k=5.

%K nonn,easy

%O 1,2

%A _Bruno Berselli_, Dec 11 2013