OFFSET
1,6
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 1.
(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, ...}.
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).
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(27) = 1 since 27 = 0 + 5 + 22 = floor(1*2/4) + floor(4*5/4) + floor(9*10/4).
a(56) = 1 since 56 = 1 + 3 + 52 = floor(2*3/4) + floor(3*4/4) + floor(14*15/4).
MATHEMATICA
S[n_]:=Union[Table[Floor[k*(k+1)/4], {k, 1, (Sqrt[16n+13]-1)/2}]]
L[n_]:=Length[S[n]]
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]];
If[Mod[Part[S[n], y], 2]==1&&MemberQ[S[n], n-Part[S[n], x]-Part[S[n], y]]==True, r=r+1];
Continue, {y, x, L[n]}]; Label[bb]; Continue, {x, 1, L[n]}]; Label[cc]; Print[n, " ", r]; Continue, {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 02 2015
STATUS
approved