cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A383359 Integers m such that m^4 is the sum of squares of two or more consecutive positive integers.

Original entry on oeis.org

13, 295, 330, 364, 1085, 5005, 6305, 15516, 415151, 1990368, 34011252, 42016497, 79565281, 139107722, 254801664, 418093065, 667378972, 1214995500, 3609736702, 4353556896
Offset: 1

Views

Author

Zhining Yang, May 01 2025

Keywords

Comments

a(21) > 10^10. - Xianwen Wang, May 08 2025
Terms of A383653 such that the consecutive integers are all positive.
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)

Examples

			295 is a term because 295^4 = 6453^2 + 6454^2 + ... + 6628^2 + 6629^2.
		

Crossrefs

Programs

  • Mathematica
    d[m_] := Select[Divisors[6 m^4], 1 < # < Floor@ CubeRoot[3 m^4] &&
        IntegerQ[1/6 (-3 (1 + #) + Sqrt[3 (12 m^4/# + 1 - #^2)])] &];
    Do[If[Length@d[m] > 0, Print[m]], {m, 1, 10000}]

Extensions

a(11)-a(18) from Xianwen Wang, May 04 2025
a(19)-a(20) from Xianwen Wang, May 08 2025

A383367 a(n) is the least integer k such that A383359(n)^4 can be expressed as a sum of squares of k consecutive integers.

Original entry on oeis.org

2, 177, 352, 1536, 2401, 40898, 60625, 185761, 19512097, 47761921, 1224370081, 7957888849, 10842382346, 11474926944, 12230369281, 190412616875, 497818686976, 72899460001, 1384334025217, 313455536641
Offset: 1

Views

Author

Zhining Yang, May 01 2025

Keywords

Comments

Terms of A383654 such that the consecutive integers are all positive.

Examples

			a(2) = 177 because A383359(2)^4 = 295^4 can be expressed as sum of squares of 177 consecutive integers: 295^4 = 6453^2 + 6454^2 + ... + 6628^2 + 6629^2.
		

Crossrefs

Extensions

a(11)-a(18) from Xianwen Wang, May 04 2025
a(19)-a(20) from Xianwen Wang, May 08 2025

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

Views

Author

Xianwen Wang, May 04 2025

Keywords

Comments

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)

Examples

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

Crossrefs

Programs

  • Mathematica
    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

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

Views

Author

Xianwen Wang, May 04 2025

Keywords

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    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
Showing 1-4 of 4 results.