A111755 Excess of n over a greedy sum of distinct squares.

0, 1, 2, 0, 0, 1, 2, 3, 0, 0, 1, 2, 0, 0, 1, 0, 0, 1, 2, 0, 0, 1, 2, 3, 0, 0, 1, 2, 0, 0, 1, 2, 3, 0, 0, 0, 0, 1, 2, 0, 0, 1, 2, 3, 0, 0, 1, 2, 0, 0, 1, 2, 0, 0, 1, 2, 3, 0, 0, 1, 2, 0, 0, 0, 0, 1, 2, 0, 0, 1, 2, 3, 0, 0, 1, 2, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 1, 2, 3, 0, 0, 1, 2, 0, 0, 1, 0, 0, 1, 0, 0, 1, 2, 0, 0

Offset: 1

Author

John W. Layman, Nov 21 2005

Keywords

• easy
• nonn

Comments

Start with the value n and subtract the largest square (not previously used) less than or equal to n to get a new value. Repeat until the value 0 is reached or the square 1 has been subtracted. The resulting value is a(n). It is not hard to prove that a(n) always lies in 0..3 inclusive.

All nonzero terms are one greater than the previous term. - Iain Fox, Oct 17 2018

Examples

			a(24)=3, since 24 -> 24 - 16 = 8 -> 8 - 4 = 4 -> 4 - 1 = 3.
		

Links

• Iain Fox, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A003995.

Programs

Mathematica

f[n_] := Block[{s = n, k = Floor@Sqrt@n}, While[k > 0, If[s >= k^2, s -= k^2]; k-- ]; s]; Array[f, 105] (* Robert G. Wilson v, Nov 22 2005 *)

PARI

a(n) = my(s=sqrtint(n)); while(s > 0, if(n >= s^2, n -= s^2); s--); n \\ Iain Fox, Oct 17 2018

Formula

a(A003995(n)) = 0. - Iain Fox, Oct 17 2018