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%I #25 Jul 01 2024 01:54:41
%S 1,2,14,6,39,11,38,17,75,62,29,117,80,88,98,165,122,59,136,207,217,
%T 231,253,265,196,297,305,323,321,329,375,385,407,411,445,447,316,483,
%U 495,513,531,535,555,561,573,583,621,651,669,675,687,705,711,735,753,767,785,789,801,819,825,855,889
%N The term that immediately precedes prime(n) in A373390.
%C In order for A373390 to contain a prime term, say a(i) = p, then there must be at least one earlier term which is a multiple of p, say a(j) = k*p with k>1 and j<i.
%C Conjectures:
%C (C1): For each prime p > 3, there is exactly one multiple of p that appears before p itself. Call this multiple k*p. Note that we know (see the Comments in A373390) that every prime appears in A373390. We will call this multiple k*p the term that "introduces" p.
%C (C2): For every prime p > 3, the introducing term k*p is always either 2*p or 3*p, and for all except the eleven primes listed in A372078 it is 2*p.
%C (C3): For every prime p > 3, the introducing term k*p occurs exactly 2 terms before p itself, with the single exception of A373390(11) = 7 which is introduced in A373390 three terms earlier, by A373390(8) = 14.
%C (C4): The primes appear in A373390 in their natural order. That is, if p<q are primes, then p appears before q. Furthermore, if k*p is the first multiple of p that appears and m*q is the first multiple of q that appears, then k*p appears before m*q.
%C Based on the limited number of known prime terms in the present sequence, i.e., 2, 11, 17, 29 and 59, it seems that for every a(n) that is prime, a(n) = A000040(n-1). - _Ivan N. Ianakiev_, Jun 22 2024
%H Michael De Vlieger, <a href="/A373790/b373790.txt">Table of n, a(n) for n = 1..40005</a> (First 6267 terms from N. J. A. Sloane)
%e A373390(24) = 11 = prime(5), so a(5) = A373390(23) = 39.
%Y Cf. A373390, A372072, A372073, A372078-A372081, A373786-A373791.
%K nonn
%O 1,2
%A _N. J. A. Sloane_, Jun 21 2024
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