A229339 GCD of all sums of n consecutive Lucas numbers.

1, 1, 2, 5, 1, 4, 1, 15, 2, 11, 1, 40, 1, 29, 2, 105, 1, 76, 1, 275, 2, 199, 1, 720, 1, 521, 2, 1885, 1, 1364, 1, 4935, 2, 3571, 1, 12920, 1, 9349, 2, 33825, 1, 24476, 1, 88555, 2, 64079, 1, 231840, 1, 167761, 2, 606965, 1, 439204, 1, 1589055, 2, 1149851, 1, 4160200, 1, 3010349, 2

Offset: 1

Author

Alonso del Arte, Sep 23 2013

Keywords

• nonn
• easy

Comments

The sum of two consecutive Lucas number is the sum of four consecutive Fibonacci numbers, which is verified easily enough with the identity L(n) = F(n - 1) + F(n + 1). Therefore a(1) = a(2) = A210209(4).

Examples

			a(3) = 2 because any sum of three consecutive Lucas numbers is an even number.
a(4) = 5 because all sums of four consecutive Lucas numbers are divisible by 5.
a(5) = 1 because some sums of five consecutive Lucas numbers are coprime.
		

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Dan Guyer and aBa Mbirika, GCD of sums of k consecutive Fibonacci, Lucas, and generalized Fibonacci numbers, Journal of Integer Sequences, 24 No.9, Article 21.9.8 (2021), 25pp; arXiv preprint, arXiv:2104.12262 [math.NT], 2021.
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,1,0,-1,0,-3,0,0,0,1).

Crossrefs

Cf. A210209, A022112, A022088, A022098, A106291 (Pisano periods of the Lucas sequence).

Programs

Mathematica

a[n_] := a[n] = If[n <= 14, {1, 1, 2, 5, 1, 4, 1, 15, 2, 11, 1, 40, 1, 29}[[n]], 3*a[n - 4] + a[n - 6] - a[n - 8] - 3*a[n - 10] + a[n - 14]]; Array[a, 64] (* Giovanni Resta, Oct 04 2013 *)
CoefficientList[Series[(x^12 - x^11 + 2 x^10 - 5 x^9 - 2 x^8 - x^7 - 6 x^6 + x^5 - 2 x^4 + 5 x^3 + 2 x^2 + x + 1) / (-x^14 + 3 x^10 + x^8 - x^6 - 3 x^4 + 1), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 09 2014 *)
LinearRecurrence[{0,0,0,3,0,1,0,-1,0,-3,0,0,0,1},{1,1,2,5,1,4,1,15,2,11,1,40,1,29},70] (* Harvey P. Dale, Jul 21 2021 *)
Table[GCD[LucasL[n + 1] - 2, LucasL[n] + 1], {n, 0, 50}] (* Horst H. Manninger, Dec 25 2021 *)

PARI

Vec(x*(x^12 -x^11 +2*x^10 -5*x^9 -2*x^8 -x^7 -6*x^6 +x^5 -2*x^4 +5*x^3 +2*x^2 +x +1) / (-x^14 +3*x^10 +x^8 -x^6 -3*x^4 +1) + O(x^100)) \\ Colin Barker, Nov 09 2014

Formula

a(n) = 3*a(n-4) + a(n-6) - a(n-8) - 3*a(n-10) + a(n-14) for n > 14. - Giovanni Resta, Oct 04 2013

G.f.: x*(x^12 -x^11 +2*x^10 -5*x^9 -2*x^8 -x^7 -6*x^6 +x^5 -2*x^4 +5*x^3 +2*x^2 +x +1) / (-x^14 +3*x^10 +x^8 -x^6 -3*x^4 +1). - Colin Barker, Nov 09 2014

From Aba Mbirika, Jan 04 2022: (Start)

a(n) = gcd(L(n+1)-1, L(n+2)-3).

a(n) = Lcm_{A106291(m) divides n} m.

Proofs of these formulas are given in Theorems 15 and 25 of the Guyer-Mbirika paper. (End)