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

1,2

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, ... .

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

Example

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.

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 *)

Crossrefs

Cf. A000330, A164576.

Sequence in context: A096564 A041196 A082669 * A146083 A321722 A306208

Adjacent sequences: A215571 A215572 A215573 * A215575 A215576 A215577

Keyword

nonn

Author

Zak Seidov, Aug 16 2012

Extensions

More terms from Alois P. Heinz, Aug 16 2012

Status

approved