A328343 Numbers k such that it is possible to find k consecutive squares whose sum is equal to the sum of two consecutive squares.

1, 2, 3, 10, 17, 25, 26, 34, 41, 50, 51, 65, 73, 82, 89, 97, 106, 113, 122, 123, 145, 146, 169, 170, 178, 185, 194, 218, 219, 241, 250, 257, 267, 274, 281, 291, 298, 305, 314, 338, 339, 353, 362, 370, 377, 386, 394, 401, 409, 410, 411, 433, 449, 457, 505, 530, 545

Offset: 1

Author

Reiner Moewald, Oct 13 2019

Keywords

• nonn

Comments

The program generates solutions to the problem, but not necessarily all solutions. To exclude the existence of further solution you have to apply coset arithmetics (modulo operations).

Examples

			k = 1: 3^2 + 4^2 = 5^2.
k = 2: 3^2 + 4^2 = 3^2 + 4^2.
k = 3: 13^2 + 14^2 = 10^2 + 11^2 + 12^2.
k = 10: 26^2 + 27^2 = 7^2 + ... + 16^2.
k = 17: 40^2 + 41^2 = 5^2 + ... + 21^2.
k = 25: 78^2 + 79^2 = 9^2 + ... + 33^2.
k = 26: 205^2 + 206^2 = 44^2 + ... + 49^2.
k = 34: 856^2 + 857^2 = 191^2 + ... + 224^2.
k = 41: 3029^2 + 3030^2 = 649^2 + ... + 689^2.
k = 50: 146^2 + 147^2 = 1^2 + ... + 50^2.
k = 51: 210^2 + 211^2 = 14^2 + ... + 64^2.
k = 65: 236^2 + 237^2 = 5^2 + ... + 69^2.
k = 73: 278^2 + 279^2 = 5^2 + ... + 76^2.
k = 82: 1070^2 + 1071^2 = 125^2 + ... + 206^2.
k = 89: 147445^2 + 147446^2 = 22059^2 + ... + 22147^2.
k = 97: 544^2 + 545^2 = 25^2 + ... + 121^2.
		

Crossrefs

Cf. A001032, A185545.

Programs

Mathematica

Select[Range[60], {} != FindInstance[ Sum[t^2, {t, x, x+#-1}] == y^2 + (y + 1)^2, {x, y}, Integers] &] (* Giovanni Resta, Oct 23 2019 *)

Python

import math
for n in range(1, 100):
   for b in range(1, 10000000):
      d = (6*b*b*(n+1)+6*b*n*(n+1)+2*n*n*n+3*n*n+n)
      w = int((math.sqrt(d/6)))
      a = w
      if 6*a*a-6*b*b*(n+1)-6*b*n*(n+1)-2*n*n*n-3*n*n-n == 0:
         print(a,b,n+1)
      a = w+1
      if 6*a*a-6*b*b*(n+1)-6*b*n*(n+1)-2*n*n*n-3*n*n-n == 0:
         print(a,b,n+1)
      a = w-1
      if 6*a*a-6*b*b*(n+1)-6*b*n*(n+1)-2*n*n*n-3*n*n-n == 0:
         print(a,b,n+1)

Extensions

a(17)-a(38) from Jon E. Schoenfield, Oct 22 2019

a(39)-a(57) from Jon E. Schoenfield and Giovanni Resta, Oct 23 2019