A210209 GCD of all sums of n consecutive Fibonacci numbers.

0, 1, 1, 2, 1, 1, 4, 1, 3, 2, 11, 1, 8, 1, 29, 2, 21, 1, 76, 1, 55, 2, 199, 1, 144, 1, 521, 2, 377, 1, 1364, 1, 987, 2, 3571, 1, 2584, 1, 9349, 2, 6765, 1, 24476, 1, 17711, 2, 64079, 1, 46368, 1, 167761, 2, 121393, 1, 439204, 1, 317811, 2, 1149851, 1, 832040

Offset

0,4

Comments

Early on in the Posamentier & Lehmann (2007) book, the fact that the sum of any ten consecutive Fibonacci numbers is a multiple of 11 is presented as an interesting property of the Fibonacci numbers. Much later in the book a proof of this fact is given, using arithmetic modulo 11. An alternative proof could demonstrate that 11*F(n + 6) = Sum_{i=n..n+9} F(i).

References

Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Prometheus Books, New York (2007) p. 33.

Links

Alois P. Heinz, Table of n, a(n) for n = 0..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.

I. D. Ruggles, Elementary problem B-1, Fibonacci Quarterly, Vol. 1, No. 1, February 1963, p. 73; Solution by Marjorie R. Bicknell, published in Vol. 1, No. 3, October 1963, pp. 76-77.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,1,0,-1,0,-3,0,0,0,1).

Formula

G.f.: -x*(x^12-x^11+2*x^10-x^9-2*x^8-x^7-6*x^6+x^5-2*x^4+x^3+2*x^2+x+1) / (x^14-3*x^10-x^8+x^6+3*x^4-1) = -1/(x^4+x^2-1) + (x^2+1)/(x^4-x^2-1) + (x+2)/(6*(x^2+x+1)) + (x-2)/(6*(x^2-x+1)) - 2/(3*(x+1)) - 2/(3*(x-1)). - Alois P. Heinz, Mar 18 2012

a(n) = gcd(Fibonacci(n+1)-1, Fibonacci(n)). - Horst H. Manninger, Dec 19 2021

From Aba Mbirika, Jan 21 2022: (Start)

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

a(n) = Lcm_{A001175(m) divides n} m.

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

Example

a(3) = 2 because all sums of three consecutive Fibonacci numbers are divisible by 2 (F(n) + F(n-1) + F(n-2) = 2F(n)), but since the GCD of 3 + 5 + 8 = 16 and 5 + 8 + 13 = 26 is 2, no number larger than 2 divides all sums of three consecutive Fibonacci numbers.
a(4) = 1 because the GCD of 1 + 1 + 2 + 3 = 7 and 1 + 2 + 3 + 5 = 11 is 1, so the sums of four consecutive Fibonacci numbers have no factors in common.

Maple

a:= n-> (Matrix(7, (i, j)-> `if`(i=j-1, 1, `if`(i=7, [1, 0, -3, -1, 1, 3, 0][j], 0)))^iquo(n, 2, 'r'). `if`(r=0, <<0, 1, 1, 4, 3, 11, 8>>, <<1, 2, 1, 1, 2, 1, 1>>))[1, 1]: seq(a(n), n=0..80);  # Alois P. Heinz, Mar 18 2012

Mathematica

Table[GCD[Fibonacci[n + 1] - 1, Fibonacci[n]], {n, 1, 50}] (* Horst H. Manninger, Dec 19 2021 *)

Prog

(PARI) a(n)=([0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0; 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0; 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0; 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0; 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0; 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0; 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1; 1, 0, 0, 0, -3, 0, -1, 0, 1, 0, 3, 0, 0, 0]^n*[0; 1; 1; 2; 1; 1; 4; 1; 3; 2; 11; 1; 8; 1])[1, 1] \\ Charles R Greathouse IV, Jun 20 2017

Crossrefs

Cf. A000045, A000071, sum of the first n Fibonacci numbers, A001175 (Pisano periods). Cf. also A229339.

Bisections give: A005013 (even part), A131534 (odd part).

Sums of m consecutive Fibonacci numbers: A055389 (m = 3, ignoring the initial 1); A000032 (m = 4, these are the Lucas numbers); A013655 (m = 5); A022087 (m = 6); A022096 (m = 7); A022379 (m = 8).

Sequence in context: A105475 A249061 A334178 * A328649 A281422 A344529

Adjacent sequences: A210206 A210207 A210208 * A210210 A210211 A210212

Keyword

nonn,easy

Author

Alonso del Arte, Mar 18 2012

Extensions

More terms from Alois P. Heinz, Mar 18 2012

Status

approved