A221471 Integers n such that n^2 is the difference of two Lucas numbers (A000032).

0, 1, 2, 3, 4, 5, 6, 11, 14, 29, 57, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371

Offset: 1

Author

T. D. Noe, Feb 13 2013

Keywords

• nonn

Comments

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.

It is easy to show that L(4n+2) - L(0) = L(2n+1)^2. - Zhi-Wei Sun, Jan 02 2015

Links

• T. D. Noe, Table of n, a(n) for n = 1..100

Crossrefs

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

Cf. A219114 (corresponding sequence for Fibonacci numbers).

Programs

Mathematica

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

Formula

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]