%I #17 Nov 22 2020 12:22:48
%S 4,1,6,1,8,1,4,1,12,5,14,1,4,1,18,1,20,1,4,1,24,1,6,1,4,1,30,21,32,1,
%T 12,1,8,1,38,1,14,1,42,1,44,1,6,1,48,1,8,1,4,1,54,1,6,9,4,1,60,1,62,1,
%U 4,1,6,1,68,1,4,1,72,1,74,1,4,1,12,1,80,1,4,1
%N a(n) is the number of consecutive terms of A100071, beginning with index n, which are divisible by n.
%C a(n) = n+1 iff n is prime.
%C a(n) = 1 iff n in { A067315 }.
%C 1 <= a(n) <= n+1.
%C { n : a(2n)>1 } = { A058008 } \ { 1 }.
%H Alois P. Heinz, <a href="/A215619/b215619.txt">Table of n, a(n) for n = 3..1000</a>
%p b:= proc(n) b(n):= n * binomial(n-1, floor((n-1)/2)) end:
%p a:= proc(n) local k;
%p for k from 0 while irem(b(n+k), n)=0 do od; k
%p end:
%p seq (a(n), n=3..100); # _Alois P. Heinz_, Aug 17 2012
%t b[n_] := n Binomial[n-1, Floor[(n-1)/2]];
%t a[n_] := Module[{k = 0}, While[Mod[b[n+k], n] == 0, k++]; k];
%t a /@ Range[3, 100] (* _Jean-François Alcover_, Nov 22 2020, after _Alois P. Heinz_ *)
%Y Cf. A000040, A067315, A100071.
%K nonn
%O 3,1
%A _Vladimir Shevelev_, Aug 17 2012