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.

A226646 Number of ways to express m/n as Egyptian fractions in just three terms, that is, m/n = 1/x + 1/y + 1/z satisfying 1 <= x <= y <= z and read by antidiagonals.

Original entry on oeis.org

3, 1, 10, 1, 3, 21, 0, 3, 8, 28, 0, 1, 3, 10, 36, 0, 1, 3, 6, 12, 57, 0, 1, 2, 3, 10, 21, 42, 0, 0, 1, 4, 2, 10, 17, 70, 0, 0, 1, 3, 3, 8, 9, 28, 79, 0, 0, 0, 1, 3, 4, 7, 20, 26, 96, 0, 0, 1, 1, 2, 3, 4, 10, 21, 36, 62, 0, 0, 0, 1, 1, 7, 1, 7, 6, 21, 25, 160, 0, 0, 0, 1, 0, 3, 3, 6, 12, 12, 16, 57, 59
Offset: 1

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Comments

See A073101 for the 4/n conjecture due to Erdös and Straus.
The first upper diagonal is 10, 8, 6, 2, 4, 1, 2, 1, 2, 0, 3, 0, 0, 1, 0, 0, 1, 0, 1, 0,... .
The main diagonal is: 3, 3, 3, 3, 3, 3, ... since 1 = 1/2 + 1/3 + 1/6 = 1/2 + 1/4 + 1/4 = 1/3 + 1/3 + 1/3. See A002966(3).
The first lower diagonal is 1, 3, 3, 4, 3, 7, 3, 5, 4, 6, 3, 10, 3, 6, 6, 6, 3, 9, 3, 9, ... .
The antidiagonal sum is 3, 11, 25, 39, 50, 79, 79, 104, 131, 157, 140, 229, 169, 220, 295, 282, ... .

Examples

			../n
m/ 1...2...3...4...5...6...7...8...9..10..11...12..13...14...15 =Allocation nbr.
.1 3..10..21..28..36..57..42..70..79..96..62..160..59..136..196 A004194
.2 1...3...8..10..12..21..17..28..26..36..25...57..20...42...81 A226641
.3 1...3...3...6..10..10...9..20..21..21..16...28..11...33...36 A226642
.4 0...1...3...3...2...8...7..10...6..12...9...21...4...17...39 A192787
.5 0...1...2...4...3...4...4...7..12..10...3...17...6...21...21 A226644
.6 0...1...1...3...3...3...1...6...8..10...7...10...1....9...12 A226645
.7 0...0...1...1...2...7...3...2...3...5...2...13...8...10....9 n/a
.8 0...0...0...1...1...3...3...3...1...2...0....8...3....7...19 n/a
.9 0...0...1...1...0...3...2...5...3...2...0....6...2....4...10 n/a
10 0...0...0...1...1...2...0...4...4...3...0....4...1....4....8 n/a
Triangle (by antidiagonals):
  {3},
  {1, 10},
  {1, 3, 21},
  {0, 3, 8, 28},
  {0, 1, 3, 10, 36},
  {0, 1, 3, 6, 12, 57},
  ...
		

Crossrefs

Programs

  • Mathematica
    f[m_, n_] := Length@ Solve[m/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Table[f[n, m - n + 1], {m, 12}, {n, m, 1, -1}] // Flatten