A094191 -id:A094191 - cp's OEIS frontend

A368041 a(n) is the least number k such that k^2 can be written as the difference of two positive squares in exactly n ways.

1, 3, 8, 16, 12, 64, 128, 24, 512, 1024, 48, 4096, 72, 60, 32768, 65536, 192, 144, 524288, 384, 2097152, 4194304, 120, 16777216, 432, 1536, 134217728, 576, 3072, 1073741824, 2147483648, 240, 1152, 17179869184, 12288, 68719476736, 137438953472, 360, 1728, 1099511627776

Offset: 0

Ilya Gutkovskiy, Dec 09 2023

nonn

Index of first occurrence of n in A046079.

All the terms are of the form 2^m * A147516(k), m >= 0, k >= 1. - Amiram Eldar, Nov 08 2024

			a(2) = 8: 8^2 = 10^2 - 6^2 = 17^2 - 15^2.

Amiram Eldar, Table of n, a(n) for n = 0..999

Cf. A025487, A046079, A068314, A071571, A094191, A100073, A122842, A147516.

a(n) = min(A122842(n+1), 2*A071571(n)). - Jon E. Schoenfield, Dec 09 2023

a(26)-a(29) from Michel Marcus, Dec 09 2023

a(30)-a(39) from Jon E. Schoenfield, Dec 09 2023

A334078 a(n) is the smallest positive integer that can be expressed as the difference of two positive squares in at least n ways.

Original entry on oeis.org

3, 15, 45, 96, 192, 240, 480, 480, 720, 960, 1440, 1440, 2880, 2880, 2880, 3360, 5040, 5040, 6720, 6720, 10080, 10080, 10080, 10080, 20160, 20160, 20160, 20160, 20160, 20160, 30240, 30240, 40320, 40320, 40320, 40320, 60480, 60480, 60480, 60480, 80640, 80640

Offset: 1

Ilya Gutkovskiy, Apr 13 2020

nonn

Eric Weisstein's World of Mathematics, Square Number

Cf. A000290, A024352, A048610, A068314, A094191, A100073, A214771, A334077.

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, ", "));

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A368041 a(n) is the least number k such that k^2 can be written as the difference of two positive squares in exactly n ways.

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A334078 a(n) is the smallest positive integer that can be expressed as the difference of two positive squares in at least n ways.

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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.

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