A355812 -id:A355812 - cp's OEIS frontend

A355813 Number of solutions (p,q) to 1/s^2 + 1/t^2 = 1/p^2 + 1/q^2 where p,q < t = 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, 3, 2, 1, 2, 2, 2, 2, 2, 2, 2, 3, 4, 4, 2, 2, 1, 2, 6, 1, 2, 6, 4, 2, 4, 1, 2, 2, 8

Offset: 1

Jianing Song, Jul 18 2022

nonn

			A355812(1) = 35. 1/s^2 + 1/35^2 = 1/p^2 + 1/q^2 has one solution, (s,p,q) = (5,7,7), so a(1) = 35.
A355812(2) = 55. 1/s^2 + 1/55^2 = 1/p^2 + 1/q^2 has two solutions, (s,p,q) = (10,11,22) and (10,22,11), so a(2) = 55.
A355812(32) = 210. 1/s^2 + 1/210^2 = 1/p^2 + 1/q^2 has three solutions, (s,p,q) = (30,42,42), (95,114,133) and (95,133,114), so a(32) = 3.

Jianing Song, Table of n, a(n) for n = 1..307

Cf. A355812, A355814.

b(n) = my(v=[;],r); for(p=1, n-1, for(q=1, n-1, r=1/(1/p^2+1/q^2-1/n^2); if(r==r\1 && issquare(r), v=concat(v,[p;q])))); v
list(lim) = my(v=[],num); for(n=1, lim, if((num=#b(n))>0, v=concat(v, num))); v

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).

Original entry on oeis.org

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), ", ")))

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).

Original entry on oeis.org

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

Jianing Song, Jan 07 2025

nonn

Partial sums of A379895.

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

			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.

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

Cf. A355812, A379895, A355813.

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, ", ")))

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.

Original entry on oeis.org

385, 425, 432, 450, 504, 585, 616, 630, 665, 693, 728, 770, 792, 800, 810, 850, 864, 900, 910, 935, 952, 1008, 1015, 1040, 1155, 1170, 1197, 1232, 1260, 1275, 1287, 1296, 1320, 1330, 1350, 1360, 1365, 1386, 1456, 1512, 1540, 1547, 1584, 1600, 1620, 1672, 1680

Offset: 1

Jianing Song, Jan 07 2025

nonn

Numbers k = A355812(r) such that A379895(r) < A355813(r).

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.

			See a-file for examples.

Jinyuan Wang, Table of n, a(n) for n = 1..2000
Mathematics Stack Exchange, Finding multiple ways of representing a number by a difference of inverse squares
Jianing Song, All examples with k <= 1500

Cf. A355812, A355813, A379895, A380150.

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

More terms from Jinyuan Wang, Jan 08 2025

A355814 Smallest value t such that 1/s^2 + 1/t^2 = 1/p^2 + 1/q^2 has exactly n solutions (p,q) where p,q < t; or -1 if no such t exists.

Original entry on oeis.org

35, 55, 210, 240, 595, 360, 560, 504, 630, 720, 1295, 1848, 1890, 1386, 1680, 2640, 2520, 3024, 5600, 3960, 2730, 4680, 6160, 8775, 9450, 5850, 5460, 5544, 9520, 15470, 5040, 7920, 9240, 25740, 10710, 9360, 13860, 13104, 8190, 17550, 10920, 18720, 15120, 22176

Offset: 1

Jianing Song, Jul 18 2022

nonn

Terms beyond a(11) = 1295 other than a(14) = 1386, if not equal to -1, are greater than 1500.

Conjecture: a(n) is divisible by 35 for odd n.

			t = 35: (s,p,q) = (5,7,7);
t = 55: (s,p,q) = (10,11,22),(10,22,11);
t = 210: (s,p,q) = (30,42,42),(95,114,133),(95,133,114);
t = 240: (s,p,q) = (70,84,112),(70,112,84),(108,135,144),(108,144,135);
t = 595: (s,p,q) = (85,91,221),(85,119,119),(85,221,91),(210,238,357),(210,357,238);
t = 360: (s,p,q) = (20,24,36),(20,36,24),(30,40,45),(30,45,40),(105,126,168),(105,168,126);
t = 560: (s,p,q) = (45,48,126),(70,80,140),(80,112,112),(45,126,48),(70,140,80),(252,315,336),(252,336,315);
t = 504: (s,p,q) = (42,56,63),(54,56,189),(42,63,56),(63,72,126),(63,126,72),(112,144,168),(112,168,144),(54,189,56);
t = 630: (s,p,q) = (35,42,63),(35,63,42),(56,63,120),(56,120,63),(90,126,126),(140,180,210),(140,210,180),(285,342,399),(285,399,342);
t = 720: (s,p,q) = (40,48,72),(40,72,48),(60,80,90),(60,90,80),(165,176,396),(210,252,336),(210,336,252),(165,396,176),(324,405,432),(324,432,405).

Cf. A328151, A355812, A355813.

b(n) = my(v=[;],r); for(p=1, n-1, for(q=1, n-1, r=1/(1/p^2+1/q^2-1/n^2); if(r==r\1 && issquare(r), v=concat(v,[p;q])))); v
search_up_to(Max,lim) = my(v=vector(Max,i,-1),num); for(n=1, lim, if((num=#b(n))>0 && num<=Max && v[num]==-1, v[num]=n)); v

a(12)-a(29) from Bert Dobbelaere, Jul 19 2022

More terms from Jinyuan Wang, Jan 25 2025

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

A355813 Number of solutions (p,q) to 1/s^2 + 1/t^2 = 1/p^2 + 1/q^2 where p,q < t = A355812(n).

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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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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).

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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.

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A355814 Smallest value t such that 1/s^2 + 1/t^2 = 1/p^2 + 1/q^2 has exactly n solutions (p,q) where p,q < t; or -1 if no such t exists.

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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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