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A379983 Numbers k such that there exists a number 1 <= m <= k-1 and at least two different pairs (x,y), 1 <= x < y <= k-1 such that 1/x^2 - 1/y^2 = 1/m^2 - 1/k^2.

%I A379983 #20 Jan 15 2025 11:14:22
%S A379983 385,425,432,450,504,585,616,630,665,693,728,770,792,800,810,850,864,
%T A379983 900,910,935,952,1008,1015,1040,1155,1170,1197,1232,1260,1275,1287,
%U A379983 1296,1320,1330,1350,1360,1365,1386,1456,1512,1540,1547,1584,1600,1620,1672,1680
%N A379983 Numbers k such that there exists a number 1 <= m <= k-1 and at least two different pairs (x,y), 1 <= x < y <= k-1 such that 1/x^2 - 1/y^2 = 1/m^2 - 1/k^2.
%C A379983 Numbers k = A355812(r) such that A379895(r) < A355813(r).
%C A379983 The smallest k such that there exists a number 1 <= m <= k-1 and at least 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, and A380150.
%H A379983 Jinyuan Wang, <a href="/A379983/b379983.txt">Table of n, a(n) for n = 1..2000</a>
%H A379983 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>
%H A379983 Jianing Song, <a href="/A379983/a379983.txt">All examples with k <= 1500</a>
%e A379983 See a-file for examples.
%o A379983 (PARI) is(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])); if(#Set(v)<#v, return(1)))); return(0) \\ See also A379895 for its program
%Y A379983 Cf. A355812, A355813, A379895, A380150.
%K A379983 nonn
%O A379983 1,1
%A A379983 _Jianing Song_, Jan 07 2025
%E A379983 More terms from _Jinyuan Wang_, Jan 08 2025