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
Keywords
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
Programs
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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;
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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
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