A221919 Triangle of numerators of sum of two unit fractions: 1/n + 1/m, n >= m >= 1.

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

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.

The column sequences give A000027(n+1), A060819(n+2), A106610(n+3), A106617(n+4), A132739(n+5), A222464 for n >= m = 1,2,..., 6.

Links

Table of n, a(n) for n=1..78.

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  (* Jean-François Alcover, Feb 25 2013 *)

Crossrefs

Cf. A221918 (companion triangle).

Sequence in context: A358106 A282510 A131225 * A341971 A273824 A137671

Adjacent sequences: A221916 A221917 A221918 * A221920 A221921 A221922

Keyword

nonn,easy,tabl,frac

Author

Wolfdieter Lang, Feb 21 2013

Status

approved