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%I #22 Aug 20 2024 03:21:42
%S 1,1,2,1,3,1,2,2,2,1,3,1,2,2,2,1,4,1,3,2,2,1,3,1,2,3,3,1,3,1,3,2,2,1,
%T 3,1,2,2,2,1,4,1,2,2,2,1,5,1,3,2,2,1,3,1,3,2,2,1,5,1,3,2,2,2,3,1,2,3,
%U 4,1,3,1,2,2,4,1,3,1,2,2,2,1,5,1,2,2,3,1,4,1,2,2,2,1,3,1,2,2,2,1,4,1,4,2,2
%N Number of k, 1 <= k <= n, such that k^3+1 divides n^3+1.
%H Alois P. Heinz, <a href="/A069929/b069929.txt">Table of n, a(n) for n = 1..10000</a>
%F Conjecture: (1/n)*Sum_{k=1..n} a(k) = C*log(log(n)) + o(log(log(n))) with 1 < C < 3/2.
%e a(5) = 3 because among the numbers 1^3+1 = 2, 2^3+1 = 9, 3^3+1 = 28, 4^3+1 = 65, and 5^3 + 1 = 126, only 3 of them (2, 9, 126) divide 5^3+1 = 126. - _Petros Hadjicostas_, Sep 18 2019
%p a:= n-> add(`if`(irem(n^3+1, k^3+1)=0, 1, 0), k=1..n):
%p seq(a(n), n=1..120); # _Alois P. Heinz_, Sep 18 2019
%o (PARI) for(n=1,150,print1(sum(i=1,n,if((n^3+1)%(i^3+1),0,1)),","))
%Y Cf. A066743.
%K easy,nonn,changed
%O 1,3
%A _Benoit Cloitre_, May 05 2002
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