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
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]
Programs
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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
Comments