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A210209
GCD of all sums of n consecutive Fibonacci numbers.
3
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
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
KEYWORD
nonn,easy
AUTHOR
Alonso del Arte, Mar 18 2012
EXTENSIONS
More terms from Alois P. Heinz, Mar 18 2012
STATUS
approved