A264105 - OEIS

%I #22 Jul 10 2020 15:11:25

%S 1,2,5,3,6,5,4,9,8,6,7,5,10,15,37,21,14,8,15,6,13,9,20,21,46,27,8,15,

%T 11,39,27,45,7,14,20,21,34,15,53,57,16,15,40,9,40,20,12,21,52,99,69,

%U 27,50,8,17,36,15,11,55,39,26,27,16,93,66,29,64,33,45,20

%N a(n) = smallest k such that n divides Sum_{i=1..k} Fibonacci(i).

%C Or smallest k such that n | A000071(k+2),

%C or smallest k such that n | A000045(k+2)-1,

%C where Fibonacci(n) = A000045(n) and A000071(n) = A000045(n)-1.

%C Remark: a(n) always exists for all n because the Fibonacci sequence is periodic mod n.

%C For all integers n, there exists an integer m such that a(m) = n. For instance if m = Fibonacci(n+2)-1, then a(m) = n.

%H Michel Lagneau, <a href="/A264105/b264105.txt">Table of n, a(n) for n = 1..1000</a>

%e a(13) = 10 because Sum_{i=1..10} Fibonacci(i) = 1+1+2+3+5+8+13+21+34+55 = 143 = 11*13 is divisible by 13. Or 13 | A000071(12) => 13|143.

%p fib:= gfun:-rectoproc({f(0)=0,f(1)=1,f(n)=f(n-1)+f(n-2)},f(n),remember):

%p a:= proc(n) local k; for k from 1 do if fib(k+2)-1 mod n = 0 then return k fi od end proc:

%p seq(a(i),i=1..1000); # _Robert Israel_, Nov 03 2015

%t Table[s=0; k=0; While[k++; s=Mod[s+Fibonacci[k], n]; s>0]; k, {n, 100}]

%t Module[{nn=100,sk},sk=Thread[{Accumulate[Fibonacci[Range[2nn]]],Range[ 2nn]}];Table[SelectFirst[sk,Divisible[#[[1]],n]&],{n,nn}]][[All,2]] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Jul 10 2020 *)

%o (PARI) a(n) = {k=1; while(k, if( (fibonacci(k+2)-1) % n == 0, return(k)); k++)} \\ _Altug Alkan_, Nov 05 2015

%Y Cf. A000045, A000071.

%K nonn

%O 1,2

%A _Michel Lagneau_, Nov 03 2015