A306597 a(n) = Card({ Sum_{k=1..n}(x_k * k) : (x_k)_{k=1..n} is an n-tuple of nonnegative integers such that Sum_{k=1..n}(x_k * T_k) = T_n }), where T_k denotes the k-th triangular number.

1, 2, 4, 6, 9, 15, 20, 27, 34, 43, 52, 63, 75, 87, 102, 117, 132, 149, 166, 185, 206, 226, 248, 271, 294, 318, 345, 373, 399, 429, 459, 489, 520, 554, 587, 623, 658, 695, 734, 772, 811, 853, 894, 936, 981, 1026, 1072, 1119, 1167, 1215, 1266, 1316, 1368, 1420

Offset

1,2

Comments

Inspired by the questions:

- Q1: into how many regions do n+1 straight lines divide the plane?

- Q2: what is the number of possible answers to Q1?

This sequence provides an answer to an analog of Q2 in a modified version of the problem.

Also an analog of A069999(n) with the roles of k and T_k swapped in the definition.

Links

Luc Rousseau, Table of n, a(n) for n = 1..150

Luc Rousseau, C program

Luc Rousseau, Java program (standalone version)

Luc Rousseau, Java program (jOEIS version, github)

Example

When n = 3, n*(n+1)/2 = 6. All possible ways to partition 6 into parts with triangular sizes (1, 3, 6) are:
  0*1 + 0*3 + 1*6 = 6
  0*1 + 2*3 + 0*6 = 6
  3*1 + 1*3 + 0*6 = 6
  6*1 + 0*3 + 0*6 = 6
In the above products, keep the left multiplicands and replace the right ones with their triangular roots:
  0*1 + 0*2 + 1*3 = 3
  0*1 + 2*2 + 0*3 = 4
  3*1 + 1*2 + 0*3 = 5
  6*1 + 0*2 + 0*3 = 6
Card({ 3, 4, 5, 6 }) = 4, so a(3) = 4.

Mathematica

T[n_] := n*(n + 1)/2
R[n_] := (Sqrt[8*n + 1] - 1)/2
S[0] := 0
S[d_] := S[d] =
  Module[{r = R[d]},
   If[IntegerQ[r], r++; r + T[r],
    First@TakeSmallest[
       1]@(S[#[[1]]] + S[#[[2]]] & /@ IntegerPartitions[d, {2}])]]
A0[n_] := Sum[Boole[d + S[d] <= 2*n], {d, 0, n}]
A[n_] := A0[T[n]]
For[n = 1, n <= 150, n++, Print[n, " ", A[n]]]

Crossrefs

Cf. A177862, A072964, A069999.

Sequence in context: A283024 A090483 A299494 * A127740 A323432 A327046

Adjacent sequences: A306594 A306595 A306596 * A306598 A306599 A306600

Keyword

nonn

Author

Luc Rousseau, Feb 27 2019

Status

approved