A365093 Smallest k such that A365092(k) = n.

2, 3, 4, 5, 7, 10, 11, 20, 22, 23, 43, 46, 47, 92, 94, 139, 188, 235, 283, 461, 517, 659, 941, 1081, 1319, 2027, 2447, 2879, 4139, 5758, 8278, 10343, 13301, 20117, 26179, 30337, 44227, 56281, 61993, 95197, 115009, 135313, 194533, 270626, 366683, 481199, 606743, 811879

Offset

2,1

Comments

a(n) is well-defined for all n >= 2.

Sequence is increasing: write a(n) as the sums and products of 1s by the process described in A365092. Removing any 1 in the expression yields a smaller number.

For m,n >= 2, it is easy to see that if gcd(a(m),a(n)) = 1, then a(m+n) <= a(m)*a(n). This is conjectured to be true for all m,n. In particular, it is conjectured that a(n+2) <= 2*a(n). If a(m+n) <= a(m)*a(n) is true, then by Fekete's subadditive lemma, we have lim_{n->oo} a(n)^(1/n) = inf_{n>=1} a(n)^(1/n) <= a(29)^(1/29) = 2879^(1/29) = 1.3160857758...

Are all terms other than 4, 20, 92 and 188 squarefree?

What are the primes that divide infinitely many terms? In particular, for p = 2 or 47, are there infinitely many terms divisible by p? Is there any term divisible by 3, 5, 7, 11 or 43 other than themselves, 235 and 517?

Links

Table of n, a(n) for n=2..49.

Jianing Song, Factorization of a(n) for n = 2..60

Example

a(9) = 20 since 20 = (1+1)^(1+1)*((1+1)^(1+1)+1) (see A365092) uses nine 1s, and all smaller numbers use fewer than nine 1s.

Prog

(PARI) A365093_search(lim) = my(list=[], k=2); for(n=1, oo, if(A365092(n)==k, list=concat(list, n); k++); if(k==lim, return(list))) \\ see A365092 for its program

Crossrefs

Cf. A365092.

Sequence in context: A005520 A048183 A255641 * A122975 A089597 A022957

Adjacent sequences: A365090 A365091 A365092 * A365094 A365095 A365096

Keyword

nonn,hard

Author

Jianing Song, Aug 21 2023

Status

approved