

A066743


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


4



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
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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



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):


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



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



