OFFSET
1,2
COMMENTS
The abundancy of k is sigma(k)/k, where sigma is the sum of divisors function, A000203; if candidates have equal abundancy, we select the smallest of them.
This sequence is motivated by a challenge to create a sequence with a concise definition that is a permutation of the positive integers and where each term differs from the next term by a single prime factor (in other words, multiply or divide by a prime). We can't specify "the lexicographically earliest such permutation": for any such permutation there is an earlier one that starts with a longer sequence of the powers of 2.
If the requirements are relaxed, some sequences derived from Gray code (A003188) serve the purpose. For example, A302033 (which generates a permutation of squarefree numbers only) and A207901 (where multiplication or division by a Fermi-Dirac factor is allowed instead of a single prime factor).
The full requirements of the challenge don't seem so easy to achieve by this approach (i.e., defining a sequence with a similarly regular pattern). In contrast, A300012 meets the challenge but lacks a simple prescription for calculating the terms, and calculating A300012 is tricky and slow.
The early terms of this sequence suggest it is a permutation. By a reasonable analysis, the appearance of a number that is a divisor of A137825(k) should usually (i.e., in more cases than not) occur before a number that is not a divisor of A137825(k+1), and numbers do not seem to appear very late if we extend this metric. (The first 23040 terms are all divisors of A137825(18) and include 20972 of 23040 divisors of A137825(17), 11313 of 11520 divisors of A137825(16), 5749 of 5760 divisors of A137825(15) and all 3840 divisors of A137825(14).)
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000
David A. Corneth, PARI program
David A. Corneth, Table of n, a(n) extended to 23040 terms as referenced in comments.
Eric Weisstein's World of Mathematics, Abundancy.
EXAMPLE
To find the next term after a(3) = 6, we first need to consider candidates of the form 6/p, for prime p. Since we need an integer, the only possible p are the prime divisors of 6, which are 2 and 3. 6/3 = 2 is already in the sequence, but 6/2 = 3 is not, so a(4) = 3, the only available candidate of the form a(3)/p.
In seeking the successor term to a(6) = 5, we see the only possible candidate of the form 5/p is 5/5 = 1, which is already in the sequence. So we consider candidates of the form 5*p. We discount p = 3 because 15 is already in the sequence, and we do not need to examine p > 5 as 5*2 will have greater abundancy than any such 5*p (since 2 and any prime p > 5 are coprime to 5, abundancy is multiplicative, and the abundancy of prime(k) decreases as k increases). The remaining candidates are 5*2 and 5*5. Of these 5*2 = 10 has the greater abundancy (1.8 compared with 1.24). So a(7) = 10.
Now we describe a more complex case: finding the next term after a(n) = a(3465) = 22496760 = 2^3 * 3^2 * 5 * 11 * 13 * 19 *23. If we divide 22496760 by any of its prime divisors, we get a number already in the sequence, disqualifying all candidates for a(n+1) of the form a(n)/p. So we consider candidates of the form a(n)*p. We discount p = 7, 17, 29 because a(n)*p is already in the sequence, and we do not need to examine p > 31 as a(n)*31 will have greater abundancy than such a(n)*p (as explained in the previous example). The remaining candidates are a(n)*p for p = 2, 3, 5, 11, 13, 19, 23, 31. Of these, the greatest abundancy is shared by a(n)*2 and a(n)*5: each has an abundancy 31/30 times the abundancy of a(n). So we choose the smaller, a(n)*2 = 44993520, to be a(n+1) = a(3466).
PROG
(PARI) \\ See Corneth link
CROSSREFS
KEYWORD
nonn
AUTHOR
David A. Corneth and Peter Munn, Apr 25 2026
STATUS
approved