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A147810 Half the number of divisors of n^2+1. 4
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 (list; graph; refs; listen; history; text; internal format)
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

LINKS

Table of n, a(n) for n=1..99.

FORMULA

a(n) = A000005(A002522(n))/2 = A147809(n)+1.

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

CROSSREFS

Cf. A048691.

Sequence in context: A251717 A057217 A193330 * A305302 A055181 A325568

Adjacent sequences:  A147807 A147808 A147809 * A147811 A147812 A147813

KEYWORD

easy,nonn

AUTHOR

M. F. Hasler, Dec 13 2008

STATUS

approved

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Last modified May 26 22:57 EDT 2019. Contains 323597 sequences. (Running on oeis4.)