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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 (list; graph; refs; listen; history; text; internal format)
 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 11F(n + 6) = sum_{i = n .. n + 9} F(i). REFERENCES Mohammad K. Azarian, Counting Sums of Nonnegative Integers, Problem H-678, Fibonacci Quarterly, Vol. 46/47, No. 4, November 2008/2009, p. 374.  Solution published in Vol. 48, No. 4, November 2010, pp. 376-377. 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 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 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 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. A000071, sum of the first n Fibonacci numbers. Cf. also A229339. Bisections give: A005013 (even part), A131534 (odd part). - Alois P. Heinz, Mar 18 2012 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 A144389 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

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Last modified August 11 00:32 EDT 2020. Contains 336403 sequences. (Running on oeis4.)