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A147810
Half the number of divisors of n^2+1.
5
1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 2, 2, 4, 1, 2, 1, 4, 3, 2, 1, 4, 2, 4, 1, 2, 1, 4, 2, 2, 2, 4, 3, 4, 2, 2, 1, 4, 3, 2, 1, 3, 2, 6, 2, 2, 2, 8, 2, 2, 2, 2, 2, 4, 1, 4, 1, 8, 2, 2, 2, 2, 2, 4, 2, 2, 1, 4, 4, 2, 3, 2, 4, 8, 1, 4, 2, 4, 2, 2, 2, 4, 3, 8, 1, 2, 2, 4, 2, 4, 1, 4, 2, 6, 1, 2, 2, 4, 4, 6
OFFSET
1,3
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) is always a positive integer.
Also number of ways to write n^2+1 as n^2+1 = x*y with 1 <= x <= y. - Michel Lagneau, Mar 10 2014
Also number of ways to write arctan(1/n) = arctan(1/x)+arctan(1/y), for integral 0 < n < x < y. - Matthijs Coster, Dec 09 2014
Number of ways that n can be expressed as (j*k-1)/(j+k) with j >= k > n. For any nonnegative integer n, the equation j*k = 1+n*(j+k) always has at least one integer solution with j >= k > n. As j >= k > n, let k=n+c (c is a positive integer), then j=n+(n^2+1)/c; we can easily conclude that c <= n, i.e., for n > 0, a(n) is the number of divisors of (n^2+1) which are <= n. - Zhining Yang, May 18 2023
FORMULA
a(n) = A000005(A002522(n))/2 = A147809(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
EXAMPLE
For n = 7 the a(7) = 3 solutions are (17,12), (32,9), (57,8). For n = 13 the a(13) = 4 solutions are (30,23), (47,18), (98,15), (183,14). - Zhining Yang, May 18 2023
MAPLE
with(numtheory); A147810:=n->tau(n^2+1)/2; seq(A147810(n), n=1..100); # Wesley Ivan Hurt, Mar 10 2014
MATHEMATICA
Table[c=0; Do[If[i<=j && i*j==n^2+1, c++], {i, t=Divisors[n^2+1]}, {j, t}]; c, {n, 100}] (* Michel Lagneau, Mar 10 2014 *)
PROG
(PARI) A147810(n)=numdiv(n^2+1)/2
(Python)
from sympy import divisor_count
def A147810(n): return divisor_count(n**2+1)>>1 if n else 1 # Chai Wah Wu, Jul 09 2023
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
M. F. Hasler, Dec 13 2008
STATUS
approved