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%I #22 Dec 16 2018 01:13:43
%S 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,
%T 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,
%U 3,1,2,1,6,25,0,6,1,6,1,1,1,6,2,25,0,1,1
%N 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.
%C Terms only depends on prime signature of n^2+1. - _David A. Corneth_, Nov 14 2018
%C 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).
%C a(A005574(n)) = 0, a(A085722(n)) = 1, a(A272078(n)) = 6, a(A316351(n)) = 25.
%C In the general case, a(k) = m if k^2+1 = p*q^m, m = 1, 2, 3, ... with p, q primes.
%C +--------------+-----------------------+------------+
%C | | number of occurrences | |
%C | a(n) | for n <= 10^5 | percentage |
%C +--------------+-----------------------+------------+
%C | 0 | 6656 | 6.656% |
%C | 1 | 23255 | 23.255% |
%C | 6 | 31947 | 31.947% |
%C | 25 | 20461 | 20.461% |
%C | other values | 17681 | 17.681% |
%C +--------------+-----------------------+------------+
%F a(n) = A089233(n^2+1). - _Michel Marcus_, Nov 13 2018
%e 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).
%p with(numtheory):nn:=10^3:
%p for n from 1 to nn do:
%p it:=0:d:=divisors(n^2+1):n0:=nops(d):
%p for k from 2 to n0-1 do:
%p for l from k+1 to n0 do:
%p if gcd(d[k],d[l])= 1
%p then
%p it:=it+1
%p else
%p fi:
%p od:
%p od:
%p printf(`%d, `,it):
%p od:
%t 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 *)
%o (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
%Y Cf. A005574, A002522, A085722, A089233, A193432, A272078, A316351.
%K nonn
%O 1,7
%A _Michel Lagneau_, Nov 12 2018
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