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

%I A365093 #20 Aug 25 2023 11:49:52
%S A365093 2,3,4,5,7,10,11,20,22,23,43,46,47,92,94,139,188,235,283,461,517,659,
%T A365093 941,1081,1319,2027,2447,2879,4139,5758,8278,10343,13301,20117,26179,
%U A365093 30337,44227,56281,61993,95197,115009,135313,194533,270626,366683,481199,606743,811879
%N A365093 Smallest k such that A365092(k) = n.
%C A365093 a(n) is well-defined for all n >= 2.
%C A365093 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.
%C A365093 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...
%C A365093 Are all terms other than 4, 20, 92 and 188 squarefree?
%C A365093 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?
%H A365093 Jianing Song, <a href="/A365093/a365093.txt">Factorization of a(n) for n = 2..60</a>
%e A365093 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.
%o A365093 (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
%Y A365093 Cf. A365092.
%K A365093 nonn,hard
%O A365093 2,1
%A A365093 _Jianing Song_, Aug 21 2023