A066743 a(n) is the number of integers of the form (n^2+1)/(k^2+1), where k = 1,2,3,...,n.

1, 1, 3, 1, 2, 1, 4, 2, 2, 1, 2, 2, 5, 1, 2, 1, 5, 3, 2, 1, 4, 2, 4, 1, 2, 1, 4, 2, 2, 2, 4, 2, 4, 1, 2, 1, 4, 3, 2, 1, 2, 2, 6, 1, 2, 1, 9, 2, 2, 1, 2, 2, 4, 1, 3, 1, 8, 2, 2, 1, 2, 2, 4, 2, 2, 1, 4, 3, 2, 1, 2, 3, 7, 1, 2, 1, 4, 2, 2, 2, 3, 2, 7, 1, 2, 1, 4, 2, 3, 1, 4, 2, 5, 1, 2, 1, 4, 3, 4, 1, 2, 2, 4

Offset

1,3

Comments

If Landau's fourth problem is ever answered in the positive, that would imply that there are infinitely many primes of the form n^2+1, in which case a(n) = 1 for infinitely many n (cf. A005574). Note that a(n) = 1 if and only if there is m >= 1 such that A066755(m) = n. - Petros Hadjicostas, Sep 18 2019

Links

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

Wikipedia, Landau's problems.

Formula

Conjecture: (1/n)*Sum_{i=1..n} a(i) = c*log(log(n)) asymptotically with 1 < c < 2.

Example

a(7) = 4 because among the numbers 1^2+1 = 2, 2^2+1 = 5, 3^2+1 = 10, 4^2+1 = 17, 5^2+1 = 26, 6^2+1 = 37, 7^2+1 = 50, exactly 4 of them (2, 5, 10, and 50) divide 7^2+1 = 50. - Petros Hadjicostas, Sep 18 2019

Maple

a:= n-> add(`if`(irem(n^2+1, k^2+1)=0, 1, 0), k=1..n):
seq(a(n), n=1..120);  # Alois P. Heinz, Sep 18 2019

Mathematica

a[ n_ ] := Length[ Select[ Range[ 1, n ], IntegerQ[ (n^2+1)/(#^2+1) ]& ] ]

Prog

(PARI) a(n) = sum(k=1, n, denominator((n^2+1)/(k^2+1)) == 1); \\ Michel Marcus, Sep 18 2019

Crossrefs

Cf. A002496, A005574, A066755, A069929 (cubic analogue).

Sequence in context: A177343 A124036 A182481 * A257915 A257904 A257980

Adjacent sequences: A066740 A066741 A066742 * A066744 A066745 A066746

Keyword

nonn

Author

Benoit Cloitre, Jan 16 2002

Extensions

Edited by Dean Hickerson, Jan 20 2002

Status

approved