login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

Half the number of divisors of n^2+1.
5

%I #45 Dec 01 2023 15:52:51

%S 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,

%T 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,

%U 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

%N Half the number of divisors of n^2+1.

%C 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.

%C Also number of ways to write n^2+1 as n^2+1 = x*y with 1 <= x <= y. - _Michel Lagneau_, Mar 10 2014

%C 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

%C 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

%H Amiram Eldar, <a href="/A147810/b147810.txt">Table of n, a(n) for n = 1..10000</a>

%H Shouen Wang, <a href="https://bbs.emath.ac.cn/forum.php?mod=redirect&amp;goto=findpost&amp;ptid=15411&amp;pid=95672&amp;fromuid=46">The general term formula of an integer sequence</a>.

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

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

%e 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

%p with(numtheory); A147810:=n->tau(n^2+1)/2; seq(A147810(n), n=1..100); # _Wesley Ivan Hurt_, Mar 10 2014

%t 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 *)

%o (PARI) A147810(n)=numdiv(n^2+1)/2

%o (Python)

%o from sympy import divisor_count

%o def A147810(n): return divisor_count(n**2+1)>>1 if n else 1 # _Chai Wah Wu_, Jul 09 2023

%Y Cf. A048691, A093582, A290332, A290333, A359225.

%K easy,nonn

%O 1,3

%A _M. F. Hasler_, Dec 13 2008