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A350178 Take n and subtract the greatest square less than or equal to n. Repeat this process until 0 is reached. a(n) is the sum of all residues after subtractions.

%I A350178 #40 Apr 27 2025 17:15:12
%S A350178 0,0,1,3,0,1,3,6,4,0,1,3,6,4,6,9,0,1,3,6,4,6,9,13,12,0,1,3,6,4,6,9,13,
%T A350178 12,9,11,0,1,3,6,4,6,9,13,12,9,11,14,18,0,1,3,6,4,6,9,13,12,9,11,14,
%U A350178 18,17,20,0,1,3,6,4,6,9,13,12,9,11,14,18,17,20,24,16,0,1,3,6,4,6,9,13
%N A350178 Take n and subtract the greatest square less than or equal to n. Repeat this process until 0 is reached. a(n) is the sum of all residues after subtractions.
%C A350178 Let s_1,s_2,s_3,...,s_m be the greedy partition of n into squares (n = s_1+s_2+s_3+...+s_m) such that s_1 >= s_2 >= s_3 >= ... >= s_m then a(n) = 0*s_1 + 1*s_2 + 2*s_3 + ... + (m-1)*s_m.
%C A350178 This sequence contains only numbers which can be written in the form c_1^2 + 2*c_2^2 + ... + m*c_m^2 with c_1 >= c_2 >= c_m. This excludes 2,5,7,8,... .
%H A350178 Thomas Scheuerle, <a href="/A350178/b350178.txt">Table of n, a(n) for n = 0..5000</a>
%F A350178 a(n) = n - r^2 + a(n - r^2) = a(n - r^2 + (b + r)^2) = a(n + b^2 + 2*b*r), r = floor(sqrt(n)), for any b >= 0. True because a(n) depends only on the distance to the next square <= n.
%F A350178 a(n) = Sum_{k>0} A053186^k(n).
%e A350178 a(41): 41 - 6^2 = 5; 5 - 2^2 = 1; 1 - 1^2 = 0 -> 5+1 = 6 = a(41).
%t A350178 Table[Total[Rest[NestWhileList[#-Floor[Sqrt[#]]^2&,n,#>0&]]],{n,0,90}] (* _Harvey P. Dale_, Apr 27 2025 *)
%o A350178 (PARI) A350178(n)={my(r=0); while(n-=sqrtint(n)^2, r+=n); r};
%Y A350178 Cf. A053186, A053610.
%K A350178 nonn,hear
%O A350178 0,4
%A A350178 _Thomas Scheuerle_, Dec 18 2021