A069929 Number of k, 1 <= k <= n, such that k^3+1 divides n^3+1.

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, 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, 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

Offset

1,3

Comments

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

Links

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Formula

Conjecture: (1/n)*Sum_{k=1..n} a(k) = C*log(log(n)) + o(log(log(n))) with 1 < C < 3/2.

a(n) < d(n^3+1). - Charles R Greathouse IV, Nov 29 2024

Example

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

Maple

a:= n-> add(`if`(irem(n^3+1, k^3+1)=0, 1, 0), k=1..n):
seq(a(n), n=1..120);  # Alois P. Heinz, Sep 18 2019

Prog

(PARI) for(n=1, 150, print1(sum(i=1, n, if((n^3+1)%(i^3+1), 0, 1)), ", "))
(PARI) a(n)=sumdiv(n^3+1, d, ispower(d-1, 3))-1 \\ Charles R Greathouse IV, Nov 30 2024

Crossrefs

Cf. A066743.

Sequence in context: A319136 A320777 A350380 * A304081 A101312 A241273

Adjacent sequences: A069926 A069927 A069928 * A069930 A069931 A069932

Keyword

easy,nonn

Author

Benoit Cloitre, May 05 2002

Status

approved