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%I #42 Oct 18 2022 16:50:12
%S 1,0,1,1,2,2,3,1,3,2,4,3,3,4,4,5,2,5,3,5,4,6,2,6,3,6,4,7,3,7,4,8,3,8,
%T 4,9,3,9,4,10,4,11,1,4,12,5,5,6,5,7,5,8,5,9,5,10,5,11,2,7,6,6,7,7,8,6,
%U 8,7,9,6,9,7,10,6,10,7,11,3,10,8,8,9,8,10
%N a(1) = 1 and for n > 0, a(n+1) is the number of k such that a(n) == a(k) (mod k) for all k = 1..n-1.
%C It is not known how fast this sequence grows or if it includes every number.
%H Joseph Bove, <a href="/A335017/b335017.txt">Table of n, a(n) for n = 1..20000</a>
%e After a(0),...,a(6) = 1, 0, 1, 1, 2, 2, a(7) = 3 because a(1) mod 1 = a(6) mod 1, a(2) mod 2 = a(6) mod 2, and a(5) mod 5 = a(6) mod 5.
%t Nest[Append[#, Count[Range[Length@ #], k_ /; Mod[#[[-1]], k] == #[[k]] ]] &, {1}, 83] (* _Michael De Vlieger_, May 24 2020 *)
%o (PARI) lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = sum(k=1, n-2, Mod(va[n-1]-va[k], k) == 0); ); va; } \\ _Michel Marcus_, May 20 2020
%K nonn
%O 1,5
%A _Joseph Bove_, May 19 2020
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