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Integers n such that n^2 is the difference of two Lucas numbers (A000032).
3

%I #11 Oct 07 2016 10:38:53

%S 0,1,2,3,4,5,6,11,14,29,57,76,199,521,1364,3571,9349,24476,64079,

%T 167761,439204,1149851,3010349,7881196,20633239,54018521,141422324,

%U 370248451,969323029,2537720636,6643838879,17393796001,45537549124,119218851371

%N Integers n such that n^2 is the difference of two Lucas numbers (A000032).

%C This sequence, growing exponentially, is interesting because the corresponding sequence for Fibonacci numbers (A219114) appears to be finite. However, except the 7 numbers in A221472, it appears that the squares of all these numbers have the form 0, L(5) - L(0), or L(4n+2) - L(0), where L(n) denotes the n-th Lucas number.

%H T. D. Noe, <a href="/A221471/b221471.txt">Table of n, a(n) for n = 1..100</a>

%F Conjecture: a(n) = 3*a(n-1)-a(n-2) = A002878(n-8) for n>13. G.f.: x^2*(28*x^11-66*x^10-16*x^9-2*x^8-13*x^7-2*x^6-5*x^5-4*x^4-3*x^3-2*x^2-x+1) / (x^2-3*x+1). [_Colin Barker_, Feb 17 2013]

%t t = Union[Flatten[Abs[Table[LucasL[n] - LucasL[i], {n, 0, 120}, {i, n}]]]]; t2 = Select[t, IntegerQ[Sqrt[#]] &]; Sqrt[t2]

%Y Cf. A000032 (Lucas numbers), A113191 (difference of two Lucas numbers).

%Y Cf. A219114 (corresponding sequence for Fibonacci numbers).

%K nonn

%O 1,3

%A _T. D. Noe_, Feb 13 2013