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

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

1,2

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)

Links

Table of n, a(n) for n=1..28.

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

Example

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

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

Crossrefs

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

Sequence in context: A371463 A346003 A164007 * A101870 A321714 A260748

Adjacent sequences: A383650 A383651 A383652 * A383654 A383655 A383656

Keyword

nonn,more

Author

Xianwen Wang, May 04 2025

Status

approved