A336186 Side length of a square block of integers, with 1 at the top-left corner, on a diagonally numbered 2D board such that the sum of the integers in the square is a perfect square.

1, 17, 127, 1871, 13969, 205793, 1536463, 22635359, 168996961, 2489683697

Offset: 1

Author

Scott R. Shannon and Eric Angelini, Jul 11 2020

Keywords

• nonn
• more

Comments

Consider a diagonally numbered 2D board shown in the example below. Draw a square, including the 1 at the top-left corner, around a block of integers and sum the integers within the square. This sequence gives the number of integers on the side of that square such that the resulting sum of integers is a perfect square.

The corresponding perfect square sum is given in A336189.

Integers m such that A185505(m) is a square. - Michel Marcus, Jul 11 2020

Examples

			Board is numbered as follows:
.
   1  2  4  7 11 16  .
   3  5  8 12 17  .
   6  9 13 18  .
  10 14 19  .
  15 20  .
  21  .
  .
a(1) = 1 is a term as 1 = 1^2 is a perfect square.
a(2) = 17 is a term as the block of integers, with the seventeen numbers {1,2,4,7,11,16,22,29,37,46,56,67,79,92,106,121,137} along the top edge and the seventeen numbers {1,3,6,10,15,21,28,36,45,55,66,78,91,105,120,136,153} along the left edge, sum to 48841 = 221^2 which is a perfect square.
		

Links

• Eric Angelini, Prime squares and square squares, personal blog "Cinquante signes", Jun. 29, 2020.
• Eric Angelini, Prime squares and square squares, personal blog "Cinquante signes", Jun. 29, 2020. [Cached copy]

Crossrefs

Cf. A336189, A185505, A000290, A000124, A000217.

Programs

PARI

isok(m) = issquare((7*m^4 + 5*m^2)/12); \\ Michel Marcus, Jul 11 2020

Formula

Conjectures from Colin Barker, Jul 11 2020: (Start)

G.f.: x*(1 + x)*(1 + 16*x + x^2) / (1 - 110*x^2 + x^4).

a(n) = 110*a(n-2) - a(n-4) for n>4. (End)

Empirical from Bill McEachen, Aug 08 2025: (Start)

a(n) = floor(k1*B^(n+1)) for odd n, and floor(k2*B^n) for even n, where k1 =(26*sqrt(21)-119)/14, k2 = (2*sqrt(21)-7)/14, and B = sqrt(55 + 12*sqrt(21)).

Above closed-forms via Amiram Eldar equate to Barker's recurrence. (End)

Extensions

a(10) from Michel Marcus, Jul 11 2020