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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.

%I A380150 #18 Jan 21 2025 03:26:50
%S A380150 2,35,385,1872,5670,30030
%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.
%C A380150 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.
%C A380150 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.
%C A380150 a(6) <= 152152, a(7) <= 318240, a(8) <= 445536, a(9) <= 1191190 (see the Mathematics Stack Exchange link).
%H A380150 Mathematics Stack Exchange, <a href="https://math.stackexchange.com/q/5019720">Finding multiple ways of representing a number by a difference of inverse squares</a>
%e A380150 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.
%o A380150 (PARI) 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
%o A380150 lista(nn) = my(k=1); for(n=0, nn, until(f(k)==n, k++); print1(k, ", "));
%Y A380150 Cf. A094191, A355812, A379983.
%K A380150 nonn,hard,more
%O A380150 0,1
%A A380150 _Jinyuan Wang_ and _Jianing Song_, Jan 13 2025