A256587 - OEIS

%I #6 Apr 03 2015 01:28:09

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

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

%U 3,4,3,5,5,4,3,5,3,6,6,5,4,7,3,6,4,7,4,8,3,7,5,6,7,6,5,6,6,6,7,8

%N Number of ways to write n = r + s + t, where r,s,t are elements of the set {floor(k*(k+1)/4): k = 1,2,3,...} with s odd and r <= s <= t.

%C Conjecture: (i) a(n) > 0 for all n > 1.

%C (ii) If the ordered pair (b,c) is among (1,2), (1,3), (1,4), (1,5), (1,6), (1,8), (1,9), (2,2) and (2,3), then each nonnegative integer can be written as r + b*s + c*t, where r,s,t belong to the set S = {floor(k*(k+1)/4): k = 1, 2, 3, ...}.

%C We have shown that if b and c are positive integers with b <= c such that every n = 0,1,2,... can be written as r + b*s + c*t with r,s,t in the above set S, then the ordered pair (b,c) must be among (1,1), (1,2,), (1,3), (1,4), (1,5), (1,6), (1,8), (1,9), (2,2) and (2,3).

%H Zhi-Wei Sun, <a href="/A256587/b256587.txt">Table of n, a(n) for n = 1..10000</a>

%e  a(27) = 1 since 27 = 0 + 5 + 22 = floor(1*2/4) + floor(4*5/4) + floor(9*10/4).

%e a(56) = 1 since 56 = 1 + 3 + 52 = floor(2*3/4) + floor(3*4/4) + floor(14*15/4).

%t S[n_]:=Union[Table[Floor[k*(k+1)/4],{k,1,(Sqrt[16n+13]-1)/2}]]

%t L[n_]:=Length[S[n]]

%t Do[r=0;Do[If[Part[S[n],x]>n/3,Goto[cc]];Do[If[Part[S[n],x]+2*Part[S[n],y]>n,Goto[bb]];

%t If[Mod[Part[S[n],y],2]==1&&MemberQ[S[n], n-Part[S[n],x]-Part[S[n],y]]==True,r=r+1];

%t Continue,{y,x,L[n]}];Label[bb];Continue,{x,1,L[n]}];Label[cc];Print[n," ",r];Continue, {n,1,100}]

%Y Cf. A000217, A256544, A256558.

%K nonn

%O 1,6

%A _Zhi-Wei Sun_, Apr 02 2015