A351532 Number of integer pairs (i, j), 0 < i, j < n, such that i/(n - i) + j/(n - j) = 1.
0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 2, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 2, 5, 0, 0, 1, 2, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 2, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 2, 0, 5, 0, 0, 1, 0, 2, 1, 0, 2, 1, 0, 0, 3, 0, 0, 1, 0, 0, 7, 0, 0, 1, 0, 0, 1, 0, 0, 1, 2, 0, 1, 0, 2, 3
Offset: 1
Keywords
Examples
For n = 3: (i, j) = (1, 1), so a(3) = 1. (1/2 + 1/2 = 1) For n = 18: (i, j) = (3, 8), (6, 6), (8, 3), so a(18) = 3. (3/15 + 8/10 = 1/5 + 4/5 = 1) For n = 20: (i, j) = (5, 8), (8, 5), so a(20) = 2. For n = 36: (i, j) = (6, 16), (8, 15), (12, 12), (15, 8), (16, 6), so a(36) = 5.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..100000
Crossrefs
Programs
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PARI
a(n)={my(x=n^2, y=2*n); sum(i=1,(n-1)/2, x-=2*n; y-=3; if(x%y==0,1,0))} -
Python
from sympy.abc import i, j from sympy.solvers.diophantine.diophantine import diop_quadratic def A351532(n): return sum(1 for d in diop_quadratic(n**2+3*i*j-2*n*(i+j)) if 0 < d[0] < n and 0 < d[1] < n) # Chai Wah Wu, Feb 15 2022
Formula
The original equation can be solved for j giving j = (n(n - 2i)) / (2n - 3i). Varying i from 1 to n-1, a(n) is given by the number of integer values of j, 0 < j < n.
Extensions
Data section extended up to a(105) by Antti Karttunen, Jan 17 2025
Comments