A152823 Largest divisor < n of n^2 + 1, a(1) = 1.

1, 1, 2, 1, 2, 1, 5, 5, 2, 1, 2, 5, 10, 1, 2, 1, 10, 13, 2, 1, 17, 5, 10, 1, 2, 1, 10, 5, 2, 17, 26, 25, 10, 13, 2, 1, 10, 17, 2, 1, 29, 5, 37, 13, 2, 29, 34, 5, 2, 41, 2, 5, 10, 1, 34, 1, 50, 5, 2, 13, 2, 5, 10, 17, 2, 1, 10, 37, 2, 29, 2, 61, 65, 1, 58, 53, 10, 5, 2, 37, 34, 25, 65, 1, 2, 13

Offset: 1

Author

M. F. Hasler, Dec 15 2008

Keywords

• nonn

Comments

If a(2k) = 3, then 4k^2 + 1 = 3p with p prime. For odd n > 1, a(n) >= 2, with equality if (n^2+1)/2 is prime. Conversely, A147809(n) = 1 iff n^2 + 1 is a semiprime, which for odd n > 1 implies a(n) = 2.

a(1) = 1 by convention, which is compatible with the FORMULA (a(n) = 1 iff n^2 + 1 is prime) and also with a(n) = the floor(d/2)-th divisor of n^2+1, when d is its total number of divisors, cf. PROGRAM. - M. F. Hasler, Sep 11 2019

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Programs

Mathematica

a[1] = 1; a[n_] := Max[Select[Divisors[n^2 + 1], # < n &]]; Array[a, 100] (* Amiram Eldar, Sep 12 2019 *)

PARI

A152823(n)={ n=divisors(n^2+1); n[ #n\2] }

Formula

a(n) = 1 iff n^2 + 1 is prime (iff A147809(n)=0), which can only happen for n = 1 or even n.