cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A280351 Expansion of Sum_{k>=0} (x/(1 - x))^(k^3).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 9, 37, 121, 331, 793, 1717, 3433, 6436, 11441, 19449, 31825, 50389, 77521, 116281, 170545, 245158, 346105, 480701, 657802, 888058, 1184419, 1564435, 2063206, 2799487, 4272049, 8544097, 23535821, 77331981, 262534537, 865287625, 2720095405
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 01 2017

Keywords

Comments

Number of compositions of n into a cube number of parts.

Examples

			a(9) = 9 because we have:
[1]  [9]
[2]  [2, 1, 1, 1, 1, 1, 1, 1]
[3]  [1, 2, 1, 1, 1, 1, 1, 1]
[4]  [1, 1, 2, 1, 1, 1, 1, 1]
[5]  [1, 1, 1, 2, 1, 1, 1, 1]
[6]  [1, 1, 1, 1, 2, 1, 1, 1]
[7]  [1, 1, 1, 1, 1, 2, 1, 1]
[8]  [1, 1, 1, 1, 1, 1, 2, 1]
[9]  [1, 1, 1, 1, 1, 1, 1, 2]
		

Crossrefs

Programs

  • Maple
    a := n -> ifelse(n = 0, 1, add(binomial(n - 1, k^3 - 1), k = 1..floor(n^(1/3)))):
    seq(a(n), n = 0..39); # Peter Luschny, Dec 23 2022
  • Mathematica
    nmax = 39; CoefficientList[Series[Sum[(x/(1 - x))^k^3, {k, 0, nmax}], {x, 0, nmax}], x]

Formula

a(0) = 1; a(n) = Sum_{k=1..floor(n^(1/3))} binomial(n-1, k^3-1) for n > 0. - Jerzy R Borysowicz, Dec 22 2022

Extensions

a(0)=1 prepended by Alois P. Heinz, Dec 17 2022

A339444 Number of partitions of n into triangular numbers such that the number of parts is a triangular number.

Original entry on oeis.org

1, 1, 0, 2, 0, 1, 2, 1, 2, 1, 4, 1, 4, 2, 3, 6, 4, 6, 4, 7, 6, 10, 8, 8, 11, 11, 14, 11, 17, 14, 22, 19, 18, 24, 24, 30, 27, 33, 31, 38, 42, 39, 47, 49, 54, 59, 60, 63, 72, 77, 79, 85, 95, 94, 104, 116, 115, 131, 133, 142, 154, 165, 168, 180, 200, 203
Offset: 0

Views

Author

Ilya Gutkovskiy, Dec 05 2020

Keywords

Examples

			                          [1]
                          [1 1]      [1]
                          [1 1 1]    [1 1]    [1]
a(10) = 4 because we have [1 1 1 1], [1 3 3], [3 6] and [10].
		

Crossrefs

Showing 1-2 of 2 results.