A215574 Minimal sum s of n distinct squares such that s is divisible by n.

1, 10, 21, 84, 55, 156, 140, 240, 342, 470, 506, 864, 819, 1176, 1395, 1616, 1785, 2214, 2470, 3260, 3570, 4092, 4324, 5184, 5525, 6578, 7101, 8456, 8555, 9750, 10416, 11712, 13134, 14314, 14910, 16776, 17575, 19570, 21099, 22760, 23821, 26166, 27434, 30096

Offset: 1

Author

Zak Seidov, Aug 16 2012

Keywords

• nonn

Comments

At n = 6k +/- 1, there is a simple formula a(n) = n*(n+1)*(2n+1)/6 and set of squares is the first n squares {1..n}^2.

Companion sequence of minimal integral averages of n distinct squares is a(n)/n: 1, 5, 7, 21, 11, 26, 20, 30, 38, 47, 46, 72, 63, 84, 93, 101, 105, 123, 130, 163, 170, 186, 188, 216, 221, ... .

Examples

			a(1) =  1 = 1^2 = 1*1.
a(2) = 10 = 1^2 + 3^2 = 2*5.
a(3) = 21 = 1^2 + 2^2 + 4^2 = 3*7.
a(4) = 84 = 1^2 + 3^2 + 5^2 + 7^2 = 4*21.
a(5) = 55 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 5*11.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..301
• Zak Seidov, Minimal sums and their n distinct summand squares for n = 1..25

Crossrefs

Cf. A000330, A164576.

Programs

Maple

b:= proc(n, i, t) option remember; local h; h:= t*(t+1)*(2*t+1)/6;
      i>=t and n>=h and (n=h or t>0 and
      (b(n, i-1, t) or i<=n and b(n-i^2, i-1, t-1)))
    end:
a:= proc(n) local s;
      for s from n*ceil((n+1)*(2*n+1)/6) by n
        while not b(s, iroot(s, 2)+1, n)
      do od; s
    end:
seq(a(n), n=1..50);  # Alois P. Heinz, Aug 16 2012

Mathematica

$RecursionLimit = 2000;
b[n_, i_, t_] := b[n, i, t] = Module[{h = t(t+1)(2t+1)/6}, i >= t && n >= h && (n==h || t>0 && (b[n, i-1, t] || i <= n && b[n-i^2, i-1, t-1]))];
a[n_] := Module[{s}, For[s = n Ceiling[(n+1)(2n+1)/6], !b[s, Floor @ Sqrt[s] + 1, n], s += n]; s];
Array[a, 50] (* Jean-François Alcover, Nov 22 2020, after Alois P. Heinz *)

Extensions

More terms from Alois P. Heinz, Aug 16 2012