A360599 Ratios of consecutive terms of A360598: a(n) = max(A360598(n), A360598(n+1)) / min(A360598(n), A360598(n+1)).

1, 2, 3, 6, 4, 5, 20, 7, 8, 56, 9, 10, 90, 11, 12, 132, 13, 14, 182, 15, 16, 240, 17, 18, 306, 19, 21, 399, 22, 23, 506, 24, 25, 600, 26, 27, 702, 28, 29, 812, 30, 31, 930, 32, 33, 1056, 34, 35, 1190, 36, 37, 1332, 38, 39, 1482, 40, 41, 1640, 42, 43, 1806, 44

Offset: 1

Author

Rémy Sigrist, Feb 13 2023

Keywords

• nonn

Comments

This sequence is a permutation of the positive integers with inverse A360600:

- we already know that all terms are distinct, so we just have to show that all integers appear,

- by contradiction: let r be the least value missing from this sequence,

- once the values 1..r-1 have appeared in this sequence, the sequence A360598 can only decrease finitely many times,

- the next increase in A360598 will correspond to the ratio r.

Examples

			For n = 15:
    A360598(15) = 11 and A360598(16) = 132,
    so a(15) = 132 / 11 = 12.
For n = 16:
    A360598(16) = 132 and A360598(17) = 1,
    so a(16) = 132 / 1 = 132.
		

Links

• Rémy Sigrist, PARI program
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A360597, A360598, A360600 (inverse).

Programs

PARI

See Links section.

Python

from itertools import islice
def agen(): # generator of terms
    an, ratios = 1, set()
    while True:
        k = 1
        q, r = divmod(max(k, an), min(k, an))
        while r != 0 or q in ratios:
            k += 1
            q, r = divmod(max(k, an), min(k, an))
        an = k
        ratios.add(q)
        yield q
print(list(islice(agen(), 66))) # Michael S. Branicky, Feb 13 2023