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%I #13 Oct 17 2018 04:40:44
%S 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,
%T 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,
%U 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
%N Excess of n over a greedy sum of distinct squares.
%C 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.
%C All nonzero terms are one greater than the previous term. - _Iain Fox_, Oct 17 2018
%H Iain Fox, <a href="/A111755/b111755.txt">Table of n, a(n) for n = 1..10000</a>
%F a(A003995(n)) = 0. - _Iain Fox_, Oct 17 2018
%e a(24)=3, since 24 -> 24 - 16 = 8 -> 8 - 4 = 4 -> 4 - 1 = 3.
%t 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 *)
%o (PARI) a(n) = my(s=sqrtint(n)); while(s > 0, if(n >= s^2, n -= s^2); s--); n \\ _Iain Fox_, Oct 17 2018
%Y Cf. A003995.
%K easy,nonn
%O 1,3
%A _John W. Layman_, Nov 21 2005
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