A256909 Enhanced triangular-number representations, concatenated.

0, 1, 2, 3, 3, 1, 3, 2, 6, 6, 1, 6, 2, 6, 3, 10, 10, 1, 10, 2, 10, 3, 10, 3, 1, 15, 15, 1, 15, 2, 15, 3, 15, 3, 1, 15, 3, 2, 21, 21, 1, 21, 2, 21, 3, 21, 3, 1, 21, 3, 2, 21, 6, 28, 28, 1, 28, 2, 28, 3, 28, 3, 1, 28, 3, 2, 28, 6, 28, 6, 1, 36, 36, 1, 36, 2

Offset

0,3

Comments

Let B = {0,1,2,3,6,10,15,21,...}, so that B consists of the triangular numbers together with 0 and 2. We call B the enhanced basis of triangular numbers. Define R(0) = 0 and R(n) = g(n) + R(n - g(n)) for n > 0, where g(n) is the greatest number in B that is <= n. Thus, each n has an enhanced triangular-number representation of the form R(n) = b(m(n)) + b(m(n-1)) + ... + b(m(k)), where b(n) > m(n-1) > ... > m(k) > 0, in which the last term, b(m(k)) is the trace.

The least n for which R(n) has more than 4 terms is given by R(7259) = 7140 + 105 + 10 + 3 + 1.

Links

Clark Kimberling, Table of n, a(n) for n = 0..1000

Example

R(0) = 0;
R(1) = 1;
R(2) = 2;
R(3) = 3;
R(4) = 3 + 1;
R(5) = 3 + 2;
R(6) = 6;
R(119) = 105 + 10 + 3 + 1.

Mathematica

b[n_] := n (n + 1)/2; bb = Insert[Table[b[n], {n, 0, 200}], 2, 3]
s[n_] := Table[b[n], {k, 1, n + 1}];
h[1] = {0, 1, 2}; h[n_] := Join[h[n - 1], s[n]];
g = h[200]; r[0] = {0};
r[n_] := If[MemberQ[bb, n], {n}, Join[{g[[n]]}, r[n - g[[n]]]]];
t = Table[r[n], {n, 0, 120}] (*A256909 before concatenation*)
Table[Last[r[n]], {n, 0, 120}]    (*A256910*)
Table[Length[r[n]], {n, 0, 120}]  (*A256911*)

Crossrefs

Cf. A000217, A256910 (trace), A256911 (number of terms), A255974 (minimal alternating triangular-number representations), A256913 (enhanced squares representations).

Sequence in context: A329315 A109199 A279813 * A343533 A308734 A279004

Adjacent sequences: A256906 A256907 A256908 * A256910 A256911 A256912

Keyword

nonn,easy

Author

Clark Kimberling, Apr 13 2015

Status

approved