A379983 -id:A379983 - cp's OEIS frontend

A379895 Number of 1 <= m <= N-1 such that there exists 1 <= x < y <= N-1 such that 1/x^2 - 1/y^2 = 1/m^2 - 1/N^2, N = A355812(n).

1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 3, 2, 2, 1, 2, 2, 2, 4, 1, 2, 4, 2, 2, 3, 2, 2, 2, 2, 2, 2, 4, 3, 2, 2, 4, 2, 1, 2, 2, 6, 2, 2, 4, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 2, 2, 1, 2, 5, 1, 2, 6, 4, 2, 4, 1, 2, 2, 7

Offset: 1

Jianing Song, Jan 05 2025

nonn

Different from A355813 at indices r such that A355812(r) is in A379983.

Let S(N) = {1/x^2 - 1/y^2 : 1 <= x < y <= N}, then N - 1 - a(n) is the size of |S(N) \ S(N-1)|, N = A355812(n). Note that S_N is the number of distinct energy differences within the first N energy levels of a hydrogen atom, and in particular |S(N)| < binomial(N,2) for N >= 35 since a(1) = 1.

			a(65) = 2 since there are 2 such m for N = A355812(65) = 385:
1/77^2 - 1/385^2 = 1/55^2 - 1/77^2 = 1/70^2 - 1/154^2;
1/154^2 - 1/385^2 = 1/70^2 - 1/77^2.
Note that A355813(65) = 3 because there are two solutions (x,y) corresponding to m = 77.
a(204) = 5 since there are 5 such m for N = A355812(204) = 1015:
1/140^2 - 1/1015^2 = 1/116^2 - 1/203^2;
1/203^2 - 1/1015^2 = 1/116^2 - 1/140^2 = 1/145^2 - 1/203^2;
1/609^2 - 1/1015^2 = 1/525^2 - 1/725^2;
1/700^2 - 1/1015^2 = 1/580^2 - 1/725^2;
1/725^2 - 1/1015^2 = 1/525^2 - 1/609^2 = 1/580^2 - 1/700^2.
Note that A355813(204) = 7 because there are two solutions (x,y) corresponding to m = 203 and to m = 725.

Jianing Song, Table of n, a(n) for n = 1..307 (corresponding to A355812(n) <= 1500)

Cf. A355812, A355813, A379983.

Partial sums give A379979.

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
for(n=1, 1500, if(b(n)>0, print1(b(n), ", ")))

A380150 a(n) is the least k such that there exists a number 1 <= m <= k-1 and exactly n different pairs (x,y), 1 <= x < y <= k-1 such that 1/x^2 - 1/y^2 = 1/m^2 - 1/k^2.

Original entry on oeis.org

2, 35, 385, 1872, 5670, 30030

Offset: 0

Jinyuan Wang and Jianing Song, Jan 13 2025

a(n) is also the least k such that there exists a number 1 <= m <= k-1 and at least n different pairs (x,y), 1 <= x < y <= k-1 such that 1/x^2 - 1/y^2 = 1/m^2 - 1/k^2: suppose on the contrary that the latter number is k' < a(n), then 1/x^2 - 1/y^2 = 1/m^2 - 1/k'^2 for some 1 <= m <= k'-1 and exactly n' > n pairs (x_1,y_1), ..., (x_{n'},y_{n'}) with 1 <= y_1 < ... < y_{n'} <= k'-1, so 1/x^2 - 1/y^2 = 1/(x_{n+1})^2 - 1/(y_{n+1})^2 has exactly n solutions (x_1,y_1), ..., (x_n,y_n) with 1 <= x < y <= y_{n+1}-1, which implies that a(n) <= y_{n+1} <= y_{n'} <= k'-1, a contradiction.
For a similar reason, this sequence is strictly increasing: if 1/x^2 - 1/y^2 = 1/m^2 - 1/a(n)^2 for some 1 <= m <= a(n)-1 and exactly n pairs (x_1,y_1), ... (x_n,y_n) with 1 <= y_1 < ... < y_n <= a(n)-1, then 1/x^2 - 1/y^2 = 1/(x_n)^2 - 1/(y_n)^2 has exactly n-1 solutions (x_1,y_1), ..., (x_{n-1},y_{n-1}) with 1 <= x < y <= y_n-1, so a(n-1) <= y_n.
a(6) <= 152152, a(7) <= 318240, a(8) <= 445536, a(9) <= 1191190 (see the Mathematics Stack Exchange link).

			The smallest k such that there exists a number 1 <= m <= k-1 and exactly three different pairs (x,y), 1 <= x < y <= k-1 such that 1/x^2 - 1/y^2 = 1/m^2 - 1/k^2 is k = 1872: we have 1/300^2 - 1/325^2 = 1/468^2 - 1/585^2 = 1/624^2 - 1/1040^2 = 1/720^2 - 1/1872^2. See the Mathematics Stack Exchange link for more examples.

Mathematics Stack Exchange, Finding multiple ways of representing a number by a difference of inverse squares

Cf. A094191, A355812, A379983.

f(k) = my(v=List([]), m2); for(y=1, k-1, for(x=1, y-1, m2=1/(1/x^2-1/y^2+1/k^2); if(m2==m2\1 && issquare(m2), listput(v, m2)))); if(#v, vecmax(vector(#v, i, sum(j=1, #v, v[i]==v[j]))), 0); \\ Gives the maximum number of pairs (x,y), 1 <= x < y <= k-1 such that 1/x^2 - 1/y^2 = 1/m^2 - 1/k^2, where m runs through 1..k-1
lista(nn) = my(k=1); for(n=0, nn, until(f(k)==n, k++); print1(k, ", "));

cp's OEIS Frontend

A379895 Number of 1 <= m <= N-1 such that there exists 1 <= x < y <= N-1 such that 1/x^2 - 1/y^2 = 1/m^2 - 1/N^2, N = A355812(n).

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