A383653 -id:A383653 - cp's OEIS frontend

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

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

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

Zhining Yang, May 01 2025

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)

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

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

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

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}]

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

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

cp's OEIS Frontend

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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