A069929 - OEIS

%I #28 Nov 30 2024 17:46:57

%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.

%C Record values are a(1) = 1, a(3) = 2, a(5) = 3, a(17) = 4, a(47) = 5, a(251) = 6, a(467) = 7, a(719) = 9, a(9299) = 10, a(30203) = 12, a(166319) = 14, a(364979) = 15, a(3080159) = 16. - _Charles R Greathouse IV_, Nov 30 2024

%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.

%F a(n) < d(n^3+1). - _Charles R Greathouse IV_, Nov 29 2024

%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)),","))

%o (PARI) a(n)=sumdiv(n^3+1,d, ispower(d-1,3))-1 \\ _Charles R Greathouse IV_, Nov 30 2024

%Y Cf. A066743.

%K easy,nonn

%O 1,3

%A _Benoit Cloitre_, May 05 2002