A147809 Half the number of proper divisors (> 1) of n^2 + 1, i.e., tau(n^2 + 1)/2 - 1.

0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 1, 1, 3, 0, 1, 0, 3, 2, 1, 0, 3, 1, 3, 0, 1, 0, 3, 1, 1, 1, 3, 2, 3, 1, 1, 0, 3, 2, 1, 0, 2, 1, 5, 1, 1, 1, 7, 1, 1, 1, 1, 1, 3, 0, 3, 0, 7, 1, 1, 1, 1, 1, 3, 1, 1, 0, 3, 3, 1, 2, 1, 3, 7, 0, 3, 1, 3, 1, 1, 1, 3, 2, 7, 0, 1, 1, 3, 1, 3, 0, 3, 1, 5, 0, 1, 1, 3, 3, 5

Offset: 1

Author

M. F. Hasler, Dec 13 2008

Keywords

• easy
• nonn

Comments

For any n > 0, n^2 + 1 cannot be a square and thus has an even number of divisors which always include 1 and n^2 + 1, therefore a(n) = (half that number minus 1) is always a nonnegative integer.

Links

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

Crossrefs

Cf. A000005, A002522, A147810, A048691, A063647, A093582.

Programs

Mathematica

DivisorSigma[0,Range[100]^2+1]/2-1 (* Harvey P. Dale, Feb 11 2015 *)

PARI

A147809(n)=numdiv(n^2+1)/2-1

Formula

a(n) = A000005(A002522(n))/2 - 1 = A147810(n) - 1.

Sum_{k=1..n} a(k) ~ c * n * log(n), where c = 3/(2*Pi) = 0.477464... (A093582). - Amiram Eldar, Dec 01 2023