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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 (list; table; graph; refs; listen; history; text; internal format)
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
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
KEYWORD
nonn,easy,tabl,frac
AUTHOR
Wolfdieter Lang, Feb 21 2013
STATUS
approved

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Last modified March 5 09:44 EST 2024. Contains 370545 sequences. (Running on oeis4.)