A256587 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.

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, 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, 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

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

Cf. A000217, A256544, A256558.

Sequence in context: A307774 A110021 A036013 * A249616 A307408 A233539

Adjacent sequences: A256584 A256585 A256586 * A256588 A256589 A256590

Keyword

nonn

Author

Zhi-Wei Sun, Apr 02 2015

Status

approved