login
A278148
Triangle T(n, m) giving in row n the numerators of the fractions for the Farey dissection of order n.
1
1, 1, 2, 1, 2, 3, 3, 1, 2, 2, 3, 5, 4, 1, 2, 2, 3, 3, 4, 5, 5, 7, 5, 1, 2, 2, 2, 3, 3, 4, 5, 5, 7, 9, 6, 1, 2, 2, 2, 3, 3, 3, 5, 4, 5, 7, 5, 7, 8, 7, 9, 11, 7, 1, 2, 2, 2, 2, 3, 3, 4, 5, 5, 4, 5, 7, 8, 7, 7, 8, 7, 9, 11, 13, 8, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 5, 5, 7, 5, 6, 9, 7, 8, 7, 7, 8, 10, 11, 9, 11, 13, 15, 9
OFFSET
1,3
COMMENTS
For the denominators see A278149.
The length of row n is A002088(n) = A005728(n) - 1.
In the Hardy reference one finds from the Farey fractions of order n >= 2 (see A006842/A006843) a dissection of the interval [1/(n+1), n/(n+1)] into A015614(n) = A005728(n) - 2 intervals J(n,j) = [l(n,j), r(n,j)], j = 1..A015614(n). They are obtained from three consecutive Farey fractions of order n: p(n,j-1)/q(n,j-1), p(n,j)/q(n,j), p(n,j+1)/q(n,j+1) by l(n,j) = p(n,j)/q(n,j) - 1/(q(n,j)*(q(n,j) + q(n,j-1))) = (p(n,j) + p(n,j-1))/(q(n,j) + q(n,j-1)) and r(n,j) = p(n,j)/q(n,j) + 1/(q(n,j)*(q(n,j) + q(n,j+1))) = (p(n,j) + p(n,j+1))/(q(n,j) + q(n,j+1)). (Hardy uses N for n, p/q - Chi_{p,q}'' for l (left) and p/q + Chi_{p,q}' for r (right)). For the second equations in l(n,j) and r(n,j) see the identities in Hardy-Wright, p. 23, Theorem 28.
Due to r(n,j) = l(n,j+1), for n >= 2 and j=1..A015614(n), it is sufficient to give for this Farey dissection of order n >= 2 only the two endpoints of the interval [1/(n+1), n/(n+1)] and the A015614(n) - 1 = A002088 - 2 = A005728(n) - 3 inner boundary points. The present table gives the numerators of these fractions. See the example section. For n = 1 we add the row 1/2 in accordance with A002088(1) = 1.
REFERENCES
G. H. Hardy, Ramanujan, AMS Chelsea Publ., Providence, RI, 2002, p. 121.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 5th ed., Clarendon Press, Oxford, 2003, pp. 23, 29 - 31.
FORMULA
T(1, 1) = 1 and for n>= 2: T(n, 1) = 1, T(n, A002088(n)) = n and for m = 2..(A002088(n) - 1): T(n, m) = numerator(l(n,m)) = numerator( p(n,m)/q(n,m) - 1/(q(n,m)*(q(n,m) + q(n,m-1)))).
EXAMPLE
The triangle T(n, m) begins:
n\m 1 2 3 4 5 6 7 8 9 10 11 12 ...
1: 1
2: 1 2
3: 1 2 3 3
4: 1 2 2 3 5 4
5: 1 2 2 3 3 4 5 5 7 5
6: 1 2 2 2 3 3 4 5 5 7 9 6
...
n = 7: 1 2 2 2 3 3 3 5 4 5 7 5 7 8 7 9 11 7,
n = 8: 1 2 2 2 2 3 3 4 5 5 4 5 7 8 7 7 8 7 9 11 13 8,
n = 9: 1 2 2 2 2 3 3 3 3 4 5 5 7 5 6 9 7 8 7 7 8 10 11 9 11 13 15 9,
n = 10: 1 2 2 2 2 2 3 3 3 5 4 4 5 5 7 5 6 9 7 8 7 9 12 8 10 11 9 11 13 15 17 10.
.............................................
The fractions T(n,m)/A278149(n, m) begin:
n\m 1 2 3 4 5 6 7 8 9 10
1: 1/2
2: 1/3 2/3
3: 1/4 2/5 3/5 3/4
4: 1/5 2/7 2/5 3/5 5/7 4/5
5: 1/6 2/9 2/7 3/8 3/7 4/7 5/8 5/7 7/9 5/6
...
n = 6: 1/7 2/11 2/9 2/7 3/8 3/7 4/7 5/8 5/7 7/9 9/11 6/7,
n = 7: 1/8 2/13 2/11 2/9 3/11 3/10 3/8 5/12 4/9 5/9 7/12 5/8 7/10 8/11 7/9 9/11 11/13 7/8,
n = 8: 1/9 2/15 2/13 2/11 2/9 3/11 3/10 4/11 5/13 5/12 4/9 5/9 7/12 8/13 7/11 7/10 8/11 7/9 9/11 11/13 13/15 8/9,
n = 9: 1/10 2/17 2/15 2/13 2/11 3/14 3/13 3/11 3/10 4/11 5/13 5/12 7/16 5/11 6/11 9/16 7/12 8/13 7/11 7/10 8/11 10/13 11/14 9/11 11/13 13/15 15/17 9/10,
n = 10: 1/11 2/19 2/17 2/15 2/13 2/11 3/14 3/13 3/11 5/17 4/13 4/11 5/13 5/12 7/16 5/11 6/11 9/16 7/12 8/13 7/11 9/13 12/17 8/11 10/13 11/14 9/11 11/13 13/15 15/17 17/19 10/11.
.............................................
For n = 5 the actual intervals J(5,j), j= 1..9 are then:
[1/6, 2/9], [2/9, 2/7], [2/7, 3/8], [3/8, 3/7], [3/7, 4/7], [4/7, 5/8], [5/8, 5/7], [5/7, 7/9], [7/9, 5/6].
KEYWORD
nonn,tabf,frac,easy
AUTHOR
Wolfdieter Lang, Nov 22 2016
STATUS
approved