This site is supported by donations to The OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A210209 GCD of all sums of n consecutive Fibonacci numbers. 3

%I

%S 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,

%T 377,1,1364,1,987,2,3571,1,2584,1,9349,2,6765,1,24476,1,17711,2,64079,

%U 1,46368,1,167761,2,121393,1,439204,1,317811,2,1149851,1,832040

%N GCD of all sums of n consecutive Fibonacci numbers.

%C 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).

%D 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.

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

%H Alois P. Heinz, <a href="/A210209/b210209.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_14">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,3,0,1,0,-1,0,-3,0,0,0,1).

%F 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

%e 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.

%e 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.

%p 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

%o (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

%Y Cf. A000071, sum of the first n Fibonacci numbers. Cf. also A229339.

%Y Bisections give: A005013 (even part), A131534 (odd part). - _Alois P. Heinz_, Mar 18 2012

%Y 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).

%K nonn,easy

%O 0,4

%A _Alonso del Arte_, Mar 18 2012

%E More terms from _Alois P. Heinz_, Mar 18 2012

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.