A379979 Number of pairs (m,k), 1 <= m < k <= N such that there exists 1 <= x < y < k-1 such that 1/x^2 - 1/y^2 = 1/m^2 - 1/k^2, N = A355812(n).
1, 3, 5, 7, 8, 10, 12, 14, 16, 17, 19, 21, 23, 25, 27, 28, 30, 32, 34, 36, 38, 40, 42, 44, 45, 47, 49, 51, 53, 55, 57, 60, 62, 64, 65, 67, 69, 71, 75, 76, 78, 82, 84, 86, 89, 91, 93, 95, 97, 99, 101, 105, 108, 110, 112, 116, 118, 119, 121, 123, 129, 131, 133, 137, 139
Offset: 1
Keywords
Examples
a(3) = 5 since A355812(3) = 56, and there are 5 such pairs (m,k), 1 <= m < k <= 56: (m,k) = (7,35): 1/5^2 - 1/7^2 = 1/7^2 - 1/35^2; (m,k) = (11,55): 1/10^2 - 1/22^2 = 1/11^2 - 1/55^2; (m,k) = (22,55): 1/10^2 - 1/11^2 = 1/22^2 - 1/55^2; (m,k) = (8,56): 1/7^2 - 1/14^2 = 1/8^2 - 1/56^2; (m,k) = (14,56): 1/7^2 - 1/8^2 = 1/14^2 - 1/56^2. Correspondingly, the set {1/x^2 - 1/y^2 : 1 <= x < y <= 56} is of size binomial(56,2) - 5.
Links
- Jianing Song, Table of n, a(n) for n = 1..307 (corresponding to A355812(n) <= 1500)
Programs
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PARI
b(n) = my(v=[], m2); for(y=1, n-1, for(x=1, y-1, m2=1/(1/x^2-1/y^2+1/n^2); if(m2==m2\1 && issquare(m2), v=concat(v, [m2])))); #Set(v) \\ #v gives A355813 my(s=0); for(n=1, 1500, if(b(n)>0, s+=b(n); print1(s, ", ")))
Comments