A178031 - OEIS

A178031

Consider the Farey tree A049455/A049456; a(n) = row at which the denominator n first appears (assumes first row is labeled row 1).

1, 2, 3, 4, 4, 6, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 7, 8, 8, 8, 8, 8, 8, 9, 8, 8, 8, 9, 9, 8, 9, 9, 9, 10, 9, 9, 9, 10, 9, 9, 9, 9, 9, 10, 9, 9, 10, 10, 10, 11, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10

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OFFSET

1,2

COMMENTS

Computed by Alan Wechsler, Dec 16, 2010.

Richard C. Schroeppel also asked about the analogous sequence giving the last occurrence of denominator n.

The first occurrence of k in this sequence is apparently at n = A135510(k-1), except for k=5. The last occurrence of k is at n = Fibonacci(k). - Andrey Zabolotskiy, Dec 01 2024

REFERENCES

Based on a posting by Richard C. Schroeppel to the Math Fun Mailing List, Dec 15 2010.

LINKS

Bo Gyu Jeong, Table of n, a(n) for n = 1..10000

Richard J. Mathar, The Kepler binary tree of reduced fractions, 2017.

EXAMPLE

Start with a pair of fractions 0/1, 1/1 and repeatedly insert the "Farey sum" (p+r)/(q+s) in between every pair of adjacent fractions p/q, r/s. The first few iterations are:

1: 0/1 1/1

2: 0/1 1/2 1/1

3: 0/1 1/3 1/2 2/3 1/1

4: 0/1 1/4 1/3 2/5 1/2 3/5 2/3 3/4 1/1

We only look at the denominators in this table (which form the sequence A049456, or A002487 if the rightmost column is removed).

1 first appears in row 1, so a(1) = 1.

2 first appears in row 2, so a(2) = 2.

3 first appears in row 3, so a(3) = 3.

4 and 5 first appear in row 4, so a(4) = a(5) = 4.

CROSSREFS

See A178047 for another version. Cf. A002487, A006842, A006843, A177903, A178042, A135510.

Sequence in context: A180046 A008329 A064558 * A008328 A298933 A365851