A383654 -id:A383654 - cp's OEIS frontend

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.

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

Offset: 1

Zhining Yang, May 01 2025

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

			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.

Cf. A001032, A189173, A383359, A383654.

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

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

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.

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