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A386855 Positive nonsquare integers of the form (r^2+s^2) / (1+r*s) for rational numbers r and s.

10, 20, 34, 52, 65, 73, 74, 130, 148, 160, 164, 202, 226, 241, 244, 265, 281, 290, 340, 394, 416, 436, 450, 452, 505, 514, 569, 577, 580, 586, 601, 641, 650, 720, 724, 745, 801, 802, 820, 848, 865, 884, 898, 916, 929, 970, 976, 1044, 1060, 1073, 1098, 1105, 1152, 1154, 1226, 1252, 1280, 1305, 1321, 1345

Offset: 1

Xianwen Wang, Aug 05 2025

nonn

We exclude perfect square cases, since Problem 6 of the 1988 International Mathematical Olympiad (IMO) proves that the expression (r^2+s^2) / (1+r*s) for integral numbers r and s yields a positive integer iff it is a perfect square.

Take a(10)=160 for example, the parametric solution is [r,s]=[(-4*U^2-296*U+23684)/(27*U^2-4320*U+27), (-788*U^2+8*U+148)/(27*U^2-4320*U+27)]

Xianwen Wang, Table of n, a(n) for n = 1..10000
Encyclopedia of Mathematics, Legendre theorem

Subsequence of A000404 and A000037.

pool=Association[];mSize=100;Block[{bc,y},Monitor[Do[bc=Table[Times@@(Select[FactorInteger[d],Mod[#[[2]],2]==1&][[All,1]]),{d,{y^2-4,y}}];If[AllTrue[bc,#>1&],If[AllTrue[{Length@Solve[x^2==bc[[1]],x,Modulus->bc[[2]]],Length@Solve[x^2==bc[[2]],x,Modulus->bc[[1]]]},#>0&],pool[y]=Lookup[pool,y,0]+1];If[Length[pool]==mSize,Break[]]],{y,1,10^10}],{y}]];Keys[pool]

A379340 Integers m such that m^2 is the sum of two or more squares of consecutive integers in more than one way.

Original entry on oeis.org

70, 105, 143, 195, 2849, 3854, 5681, 8075, 143737, 144157, 208395, 939356, 1226670, 2259257, 2656724, 2741046, 4598528, 6555549, 7832413, 11818136, 19751043, 32938290, 429323037, 807759678, 1375704770, 1656510196, 1981351834

Offset: 1

Xianwen Wang, May 23 2025

nonn

			105^2 = (-19)^2 + (-18)^2 + ... + 29^2 = (-21)^2 + (-20)^2 + ... + 28^2.
143^2 = 38^2 + 39^2 + ... + 48^2 = 7^2 + 8^2 + ... + 39^2.
2259257^2 = 26181^2 + 26182^2 + ... + 32158^2 = 9401^2 + 9402^2 + ... + 25273^2.

Xianwen Wang, Rust program

Cf. A062681.

Subsequence of A174069.

A384440 Array of triples (x,y,z) of minimal (positive) solutions of the cubic Pell equation x^3 + ny^3 + n^2z^3 - 3nxyz = 1, read by rows.

Original entry on oeis.org

1, 0, 0, 1, 1, 1, 4, 3, 2, 5, 3, 2, 41, 24, 14, 109, 60, 33, 4, 2, 1, 1, 0, 0, 4, 2, 1, 181, 84, 39, 89, 40, 18, 9073, 3963, 1731, 94, 40, 17, 29, 12, 5, 5401, 2190, 888, 16001, 6350, 2520, 324, 126, 49, 55, 21, 8, 64, 24, 9, 361, 133, 49

Offset: 1

Xianwen Wang, May 29 2025

Given n, n!=k^3, there are infinitely many solutions, and all other solutions can be derived from the minimal solution pair by a recurrence relation. See Wolfe, pages 359-369.

			For n=5, the minimal positive solution is (41, 24, 14), so a(13)=41, a(14)=24, a(15)=14.
The array begins:
    1,  0,  0,
    1,  1,  1,
    4,  3,  2,
    5,  3,  2,
   41, 24, 14,
  109, 60, 33,
  ...

Clyde Lynne Earle Wolfe, On the Indeterminate Cubic Equation X^3 + Dy^3 + D^2z^3 - 3Dxyz, University of California Press, 1923, pp. 359-369.

Xianwen Wang, Table of n, a(n) for n = 1..6000

Name edited by Michel Marcus, Jun 03 2025

A383654 a(n) is the number k such that A383653(n)^4 is the sum of squares of k consecutive integers.

Original entry on oeis.org

2, 2, 169, 242, 177, 352, 1536, 2401, 40898, 163607, 230121, 60625, 218089, 185761, 19512097, 47761921, 1170329056, 1224370081, 7957888849, 10842382346, 11474926944, 208152552417, 12230369281, 190412616875, 497818686976, 72899460001, 1384334025217, 313455536641

Offset: 1

Xianwen Wang, May 04 2025

			Case a(1)=2: 13^4 = 119^2 + 120^2, 1^4 = 0^2 + 1^2.
Case a(3)=169: 26^4 = (-67+1)^2 + (-67+2)^2 + ... + (-67+168)^2 + (-67+169)^2.
Case a(5)=177: 295^4 = (6452+1)^2 + (6452+2)^2 + ... + (6452+176)^2 + (6452+177)^2.
...
Case a(10)=163607: 5546^4 = (-22206+1)^2 + (-22206+2)^2 + ... + (-22206+163606)^2 + (-22206+163607)^2.

Zhining Yang, Can be expressed as the fourth power of the sum of squares of consecutive positive integers, Chinese BBS.

Cf. A000330, A097812, A189173, A383359, A383367, A383653.

lst={};Monitor[Do[mm=6 m^4;div=TakeWhile[Divisors[mm][[2;;-2]],2mm/#+1>#^2&];
ans=Select[div,IntegerQ[Sqrt[(2mm/#+1-#^2)/3]]&&Mod[#-Sqrt[(2mm/#+1-#^2)/3],2]==1&];
If[Length[ans]>0,tmp={m,{#,q=Sqrt[(2mm/#+1-#^2)/3],p=(q+1-#)/2}&/@ans};Print[tmp];
AppendTo[lst,tmp]],{m,1,10^4}],m];lst

A383653 Integers m such that m^4 is the sum of squares of two or more consecutive integers, positive or negative.

Original entry on oeis.org

1, 13, 26, 33, 295, 330, 364, 1085, 5005, 5546, 5682, 6305, 6538, 15516, 415151, 1990368, 3538366, 34011252, 42016497, 79565281, 139107722, 175761059, 254801664, 418093065, 667378972, 1214995500, 3609736702, 4353556896

Offset: 1

Xianwen Wang, May 04 2025

a(29) > 10^10.
From David A. Corneth, May 04 2025: (Start)
The sum of the first m positive squares is f(m) = m*(m + 1)*(2*m + 1) / 6.
The sum of consecutive squares m^2 + (m+1)^2 + ... + t^2 where 0 < m <= t may be written as f(t) - f(m-1) for some t and m.
From there we can factor out t - m - 1 and solve the system of equations going over divisors of 6*m^4.
To get divisors of 6*m^4 we need to factor 6*m^4 which can be done using the factors of 6 and the factors of m. Doing so makes we need to factorize smaller numbers. (End)

			5546 is a term because 5546^4 = (-22205)^2 + (-22204)^2 + ... + 141400^2 + 141401^2.

Zhining Yang, Can be expressed as the fourth power of the sum of squares of consecutive positive integers, Chinese BBS.

Cf. A000330, A097812, A189173, A383359, A383367, A383654.

lst={};Monitor[Do[mm=6 m^4;div=TakeWhile[Divisors[mm][[2;;-2]],2mm/#+1>#^2&];
ans=Select[div,IntegerQ[Sqrt[(2mm/#+1-#^2)/3]]&&Mod[#-Sqrt[(2mm/#+1-#^2)/3],2]==1&];
If[Length[ans]>0,tmp={m,{#,q=Sqrt[(2mm/#+1-#^2)/3],p=(q+1-#)/2}&/@ans};Print[tmp];
AppendTo[lst,tmp]],{m,1,10^4}],m];lst

cp's OEIS Frontend

User: Xianwen Wang

A386855 Positive nonsquare integers of the form (r^2+s^2) / (1+r*s) for rational numbers r and s.

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A379340 Integers m such that m^2 is the sum of two or more squares of consecutive integers in more than one way.

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A384440 Array of triples (x,y,z) of minimal (positive) solutions of the cubic Pell equation x^3 + n*y^3 + n^2*z^3 - 3*n*x*y*z = 1, read by rows.

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A383654 a(n) is the number k such that A383653(n)^4 is the sum of squares of k consecutive integers.

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Author

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Programs

Mathematica

A383653 Integers m such that m^4 is the sum of squares of two or more consecutive integers, positive or negative.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A384440 Array of triples (x,y,z) of minimal (positive) solutions of the cubic Pell equation x^3 + ny^3 + n^2z^3 - 3nxyz = 1, read by rows.