OFFSET
1,2
COMMENTS
Interestingly, the minimum value of a(n) - a(n-1) is 1. Is there a maximum value of a(n) - a(n-1)?
From Robert Israel, Oct 19 2015: (Start)
n is in the sequence if either n is odd and A001175(n) and A001175((n+1)/2) both divide n+1, or n is even and A001175(n/2) and A001175(n+1) both divide n.
Most of the terms of the sequence appear to fall in these categories. The first two that do not are 15456 and 41640.
In particular, if n = 2^j * 3^k * 5^m with j >= 4, k >= 1 and m >= 1, and n+1 is prime, then n is in the sequence. There are believed to be infinitely many numbers of this form. The first few are 240, 1200, 2160, 4800, 6480, 7680, 8640, 9600, 14400, 15360. (End)
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
MAPLE
fmod:= proc(a, b) local A, t;
uses LinearAlgebra[Modular];
if b < 4295022903 then t:= integer[8] else t:= integer fi;
A:= Mod(b, <<1, 1>|<1, 0>>, t);
MatrixPower(b, A, a)[1, 2];
end proc:
filter:= n -> (fmod(n+2, n*(n+1)/2) = 1):
filter(1):= true:
select(filter, [$1..10^5]); # Robert Israel, Oct 19 2015
MATHEMATICA
fQ[n_] := Mod[Fibonacci[n + 2] - 1, n (n + 1)/2] == 0; Select[Range@20000, fQ] (* Bruno Berselli, Oct 19 2015 - after Robert G. Wilson v in A263225 *)
PROG
(PARI) for(n=1, 20000, if((fibonacci(n+2)-1) % (n*(n+1)/2) == 0, print1(n", ")));
(PARI) is(n)=((Mod([1, 1; 1, 0], n*(n+1)/2))^(n+2))[1, 2]==1 \\ Charles R Greathouse IV, Oct 19 2015
(Magma) [n: n in [1..20000] | IsDivisibleBy(Fibonacci(n+2)-1, n*(n+1) div 2)]; // Bruno Berselli, Oct 19 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Altug Alkan, Oct 11 2015
STATUS
approved