

A221919


Triangle of numerators of sum of two unit fractions: 1/n + 1/m, n >= m >= 1.


4



2, 3, 1, 4, 5, 2, 5, 3, 7, 1, 6, 7, 8, 9, 2, 7, 2, 1, 5, 11, 1, 8, 9, 10, 11, 12, 13, 2, 9, 5, 11, 3, 13, 7, 15, 1, 10, 11, 4, 13, 14, 5, 16, 17, 2, 11, 3, 13, 7, 3, 4, 17, 9, 19, 1, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 2, 13, 7, 5, 1, 17, 1, 19, 5, 7, 11, 23, 1
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OFFSET

1,1


COMMENTS

The triangle of the corresponding denominators is given in A221918.
See A221918 for comments on resistance, reduced mass and radius of the twin circles in Archimedes's arbelos, as well as references.


LINKS



FORMULA

a(n,m) = numerator(2/n + 1/m), n >= m >= 1, and 0 otherwise.
A221918(n,m)/a(n,m) = R(n,m) = n*m/(n+m). 1/R(n,m) = 1/n + 1/m.


EXAMPLE

The triangle a(n,m) begins:
n\m 1 2 3 4 5 6 7 8 9 10 11 12 ...
1: 2
2: 3 1
3: 4 5 2
4: 5 3 7 1
5: 6 7 8 9 2
6: 7 2 1 5 11 1
7: 8 9 10 11 12 13 2
8: 9 5 11 3 13 7 15 1
9: 10 11 4 13 14 5 16 17
10: 11 3 13 7 3 4 17 9 19 1
11: 12 13 14 15 16 17 18 15 20 21 2
12: 13 7 5 1 17 1 19 5 7 11 23 1
...
a(n,1) = n + 1 because R(n,1) = n/(n+1), gcd(n,n+1) = 1, hence denominator(R(n,m)) = n + 1.
a(5,4) = 9 because R(5,4) = 20/9, gcd(20,9) = 1, hence denominator( R(5,4)) = 9.
a(6,3) = 1 because R(6,3) = 18/9 = 2/1.
For the rationals R(n,m) see A221918.


MATHEMATICA

a[n_, m_] := Numerator[1/n + 1/m]; Table[a[n, m], {n, 1, 12}, {m, 1, n}] // Flatten (* JeanFrançois Alcover, Feb 25 2013 *)


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



