A321519 Let d(n,i), i = 1..k be the k divisors of n^2 + 1 (the number 1 is not counted). a(n) is the number of ordered pairs d(n,i) < d(n,j) such that gcd(d(n,i), d(n,j)) = 1.

0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 1, 1, 6, 0, 1, 0, 6, 2, 1, 0, 6, 1, 6, 0, 1, 0, 6, 1, 1, 1, 6, 2, 6, 1, 1, 0, 6, 2, 1, 0, 2, 1, 11, 1, 1, 1, 25, 1, 1, 1, 1, 1, 6, 0, 6, 0, 16, 1, 1, 1, 1, 1, 6, 1, 1, 0, 6, 3, 1, 2, 1, 6, 25, 0, 6, 1, 6, 1, 1, 1, 6, 2, 25, 0, 1, 1

Offset: 1

Author

Michel Lagneau, Nov 12 2018

Keywords

• nonn

Comments

Terms only depends on prime signature of n^2+1. - David A. Corneth, Nov 14 2018

We observe an interesting statistic for n <= 10^5: the four values of a(n) = 0, 1, 6, 25 represent more than 82% (see the table below).

a(A005574(n)) = 0, a(A085722(n)) = 1, a(A272078(n)) = 6, a(A316351(n)) = 25.

In the general case, a(k) = m if k^2+1 = p*q^m, m = 1, 2, 3, ... with p, q primes.

+--------------+-----------------------+------------+

| | number of occurrences | |

| a(n) | for n <= 10^5 | percentage |

+--------------+-----------------------+------------+

| 0 | 6656 | 6.656% |

| 1 | 23255 | 23.255% |

| 6 | 31947 | 31.947% |

| 25 | 20461 | 20.461% |

| other values | 17681 | 17.681% |

+--------------+-----------------------+------------+

Examples

			a(13) = 6 because the divisors {d(i)} of 13^2 + 1 = 170 (without the number 1)  are  {2, 5, 10, 17, 34, 85, 170}, and gcd(d(i), d(j)) = 1 for the 6 following pairs of elements of {d(i)}: (2, 5), (2, 17), (2, 85), (5, 17), (5, 34) and (10, 17).
		

Crossrefs

Cf. A005574, A002522, A085722, A089233, A193432, A272078, A316351.

Programs

Maple

with(numtheory):nn:=10^3:
for n from 1 to nn do:
  it:=0:d:=divisors(n^2+1):n0:=nops(d):
   for k from 2 to n0-1 do:
    for l from k+1 to n0 do:
     if gcd(d[k],d[l])= 1
      then
      it:=it+1
      else
     fi:
   od:
  od:
  printf(`%d, `,it):
od:

Mathematica

f[n_] := (DivisorSigma[0, n^2] - 1)/2 - DivisorSigma[0, n] + 1; Map[f, Range[0,100]^2+1] (* Amiram Eldar, Nov 14 2018 after Robert G. Wilson v at A089233 *)

PARI

a(n) = {my(d=divisors(n^2+1)); sum(k=2, #d, sum(j=2, k-1, gcd(d[k], d[j]) == 1));} \\ Michel Marcus, Nov 12 2018

Formula

a(n) = A089233(n^2+1). - Michel Marcus, Nov 13 2018