cp's OEIS Frontend

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.

%I A226646 #28 Feb 16 2025 08:33:20
%S A226646 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,
%T A226646 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,
%U A226646 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
%N 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.
%C A226646 See A073101 for the 4/n conjecture due to Erdös and Straus.
%C A226646 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,... .
%C A226646 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).
%C A226646 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, ... .
%C A226646 The antidiagonal sum is 3, 11, 25, 39, 50, 79, 79, 104, 131, 157, 140, 229, 169, 220, 295, 282, ... .
%H A226646 Christian Elsholtz, <a href="http://www.ams.org/tran/2001-353-08/S0002-9947-01-02782-9/home.html">Sums Of k Unit Fractions</a>
%H A226646 David Eppstein, <a href="http://www.ics.uci.edu/~eppstein/numth/egypt/intro.html">Algorithms for Egyptian Fractions</a>
%H A226646 David Eppstein, <a href="http://library.wolfram.com/infocenter/Articles/2926/">Ten Algorithms for Egyptian Fractions</a>
%H A226646 Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html">Egyptian Fractions</a>
%H A226646 Oakland University, <a href="http://www.oakland.edu/enp/">The Erdős Number Project</a>
%H A226646 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/EgyptianFraction.html">Egyptian Fraction</a>
%H A226646 <a href="/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>
%e A226646 ../n
%e A226646 m/ 1...2...3...4...5...6...7...8...9..10..11...12..13...14...15 =Allocation nbr.
%e A226646 .1 3..10..21..28..36..57..42..70..79..96..62..160..59..136..196 A004194
%e A226646 .2 1...3...8..10..12..21..17..28..26..36..25...57..20...42...81 A226641
%e A226646 .3 1...3...3...6..10..10...9..20..21..21..16...28..11...33...36 A226642
%e A226646 .4 0...1...3...3...2...8...7..10...6..12...9...21...4...17...39 A192787
%e A226646 .5 0...1...2...4...3...4...4...7..12..10...3...17...6...21...21 A226644
%e A226646 .6 0...1...1...3...3...3...1...6...8..10...7...10...1....9...12 A226645
%e A226646 .7 0...0...1...1...2...7...3...2...3...5...2...13...8...10....9 n/a
%e A226646 .8 0...0...0...1...1...3...3...3...1...2...0....8...3....7...19 n/a
%e A226646 .9 0...0...1...1...0...3...2...5...3...2...0....6...2....4...10 n/a
%e A226646 10 0...0...0...1...1...2...0...4...4...3...0....4...1....4....8 n/a
%e A226646 Triangle (by antidiagonals):
%e A226646   {3},
%e A226646   {1, 10},
%e A226646   {1, 3, 21},
%e A226646   {0, 3, 8, 28},
%e A226646   {0, 1, 3, 10, 36},
%e A226646   {0, 1, 3, 6, 12, 57},
%e A226646   ...
%t A226646 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
%Y A226646 Cf. A227612, A226640, A226641, A226642, A192787, A226644, A226645.
%K A226646 nonn,tabl
%O A226646 1,1
%A A226646 _Allan C. Wechsler_ and _Robert G. Wilson v_, Aug 17 2013