A376202 Number of pairs 1 <= x <= y <= n-1 such that gcd(x,n) = gcd(y,n) = gcd(x+y,n) = 1 and 1/x + 1/y == 1/(x+y) mod n.

0, 0, 2, 0, 0, 0, 6, 0, 0, 0, 0, 0, 12, 0, 0, 0, 0, 0, 18, 0, 12, 0, 0, 0, 0, 0, 0, 0, 0, 0, 30, 0, 0, 0, 0, 0, 36, 0, 24, 0, 0, 0, 42, 0, 0, 0, 0, 0, 42, 0, 0, 0, 0, 0, 0, 0, 36, 0, 0, 0, 60, 0, 0, 0, 0, 0, 66, 0, 0, 0, 0, 0, 72, 0, 0, 0, 0, 0, 78, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 144, 0, 60, 0, 0, 0, 96, 0, 0, 0, 0, 0, 102, 0, 0

Offset: 1

Author

Tom Duff and N. J. A. Sloane, Oct 06 2024

Keywords

• nonn

Comments

In general, 1/x + 1/y = 1/(x+y) is the wrong way to add fractions!

See A376203 for a(2*n-1)/2 and A376755 for a(6*n+1)/6.

From Robert Israel, Nov 06 2024: (Start)

If a(n) = 0 then a(m) = 0 whenever m is a multiple of n.

It appears that the primes p for which a(p) > 0 are A007645. (End)

Examples

			For n = 3 the a(3) = 2 solutions are (x,y) = (1,1) and (2,2).
For n = 7 the a(7) = 6 solutions are (x,y) = (1,2), (1,4), (2,4), (3,5), (3,6), (5,6).
		

Links

• Robert Israel, Table of n, a(n) for n = 1..6000

Crossrefs

Cf. A000086, A007645, A046530, A290731, A376203, A376755, A376756, A376757.

Programs

Maple

a:=[];
for n from 1 to 140 do
c:=0;
for y from 1 to n-1 do
for x from 1 to y do
if gcd(y,n) = 1 and gcd(x,n) = 1 and gcd(x+y,n) = 1  and (1/x + 1/y - 1/(x+y)) mod n = 0 then c:=c+1; fi;
od: # od x
od: # od y
a:=[op(a),c];
od: # od n
a;

Python

from math import gcd
def A376202(n):
    c = 0
    for x in range(1,n):
        if gcd(x,n) == 1:
            for y in range(x,n):
                if gcd(y,n)==gcd(z:=x+y,n)==1 and not (w:=z**2-x*y)//gcd(w,x*y*z)%n:
                    c += 1
    return c # Chai Wah Wu, Oct 06 2024