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a(n) = n-floor(sqrt(n))^2-floor(sqrt(n-floor(sqrt(n))^2))^2; difference between n and sum of two largest distinct squares <= n.
2

%I #24 Dec 17 2022 06:43:19

%S 0,0,0,1,0,0,1,2,0,0,0,1,2,0,1,2,0,0,1,2,0,1,2,3,4,0,0,1,2,0,1,2,3,4,

%T 0,1,0,0,1,2,0,1,2,3,4,0,1,2,3,0,0,1,2,0,1,2,3,4,0,1,2,3,4,5,0,0,1,2,

%U 0,1,2,3,4,0,1,2,3,4,5,6,0,0,0,1,2,0,1,2,3,4,0,1,2,3,4,5,6,0,1,2,0,0,1,2,0

%N a(n) = n-floor(sqrt(n))^2-floor(sqrt(n-floor(sqrt(n))^2))^2; difference between n and sum of two largest distinct squares <= n.

%C If a(n)=0 then n is a sum of two squares A001481, but not conversely. For the sum of two squares n = 18, 32, 41, ... we have a(n) > 0. - _Thomas Ordowski_, Jul 11 2014

%H Michel Marcus, <a href="/A174806/b174806.txt">Table of n, a(n) for n = 0..10000</a>

%F a(n) = 0 iff A053610(n) < 3 and 0 < a(n) = m^2 iff A053610(n) = 3. - _Thomas Ordowski_, Jul 12 2014

%e 24=4^2+8;8-2^2=4, 115=10^2+15;15-3^2=6,..

%t a[n_]:=n-Floor[Sqrt[n]]^2-Floor[Sqrt[n-Floor[Sqrt[n]]^2]]^2;

%t Table[a[n], {n,0,6!}]

%o (PARI) a(n) = my(x=sqrtint(n)^2); n - x - sqrtint((n-x))^2; \\ _Michel Marcus_, Dec 17 2022

%Y Cf. A001481, A053186, A053610.

%K nonn

%O 0,8

%A _Vladimir Joseph Stephan Orlovsky_, Mar 29 2010