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A147809
Half the number of proper divisors (> 1) of n^2 + 1, i.e., tau(n^2 + 1)/2 - 1.
6
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
1,7
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
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
MATHEMATICA
DivisorSigma[0, Range[100]^2+1]/2-1 (* Harvey P. Dale, Feb 11 2015 *)
PROG
(PARI) A147809(n)=numdiv(n^2+1)/2-1
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
M. F. Hasler, Dec 13 2008
STATUS
approved