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%I #24 Aug 09 2022 07:07:56
%S 1,5,6,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,24,25,26,27,28,29,
%T 30,31,32,33,34,35,36,37,38,39,40,41,42,44,45,46,47,48,49,50,51,52,53,
%U 54,55,56,57,58,59,60,61,62,63,64,65,66,68,69,70,71,72,73,74,75,76,77,78,79,80
%N Integers k such that A003266(k) is divisible by k.
%C Note that this sequence is not the complement of A000057.
%C See A230359 for the prime terms of this sequence.
%H Amiram Eldar, <a href="/A270653/b270653.txt">Table of n, a(n) for n = 1..10000</a>
%e 11 is a term because 1*1*2*3*5*8*13*21*34*55*89 is divisible by 11.
%t Select[Range@ 80, Divisible[Fibonorial@ #, #] &] (* Version 10, or *) Select[Range@ 80, Divisible[Product[Fibonacci@ k, {k, #}], #] &] (* _Michael De Vlieger_, Mar 23 2016 *)
%o (PARI) t(n) = prod(k=1, n, Mod(fibonacci(k), n));
%o for(n=1, 1e2, if(lift(t(n)) == 0, print1(n, ", ")));
%Y Complement of A270777.
%Y Cf. A000057, A003266, A230359.
%K nonn
%O 1,2
%A _Altug Alkan_, Mar 20 2016
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