A256910 -id:A256910 - cp's OEIS frontend

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

Clark Kimberling, Apr 13 2015

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.

			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.

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

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

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*)

A256911 Number of terms in the enhanced triangular-number representation of n.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 1, 2, 2, 2, 3, 3, 1, 2, 2, 2, 3, 3, 2, 1, 2, 2, 2, 3, 3, 2, 3, 1, 2, 2, 2, 3, 3, 2, 3, 3, 1, 2, 2, 2, 3, 3, 2, 3, 3, 3, 1, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 1, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 1, 2, 2, 2, 3, 3, 2, 3

Offset: 0

Clark Kimberling, Apr 14 2015

See A256909 for definitions. The least n for which R(n) has 5 terms is given by R(7259) = 7140 + 105 + 10 + 3 + 1.

			R(4) = 3 + 1, so a(4) = 2.
R(119) = 105 + 10 + 3 + 1, so a(119) = 4.

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

Cf. A000217, A256909 (representations) A256910 (trace).

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*)
Flatten[t]  (*A256909*)
Table[Last[r[n]], {n, 0, 120}]    (*A256910*)
Table[Length[r[n]], {n, 0, 120}]  (*A256911*)

cp's OEIS Frontend

A256909 Enhanced triangular-number representations, concatenated.

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