%I #21 Sep 04 2023 12:45:16
%S 12,12,30,63,154,273,442,646,874,1334,1798,2294,3034,3526,4042,4982,
%T 6254,7198,8174,9514,10366,11534,13114,14774,17266,19594,20806,22042,
%U 23326,24634,28702,33274,35894,38086,41422,44998,47414,51182,54442,57782,61934,64798,69142
%N a(1) = 12; for n >= 2, a(n) = least positive integer of the form prime(m)*prime(nm)*prime(n) with m >= 1.
%C Also a(n) is the least positive integer in A364462 that is divisible by prime(n).
%C This sequence is strictly increasing for n > 1.
%C Proof by contradiction:
%C Suppose a(n) >= a(n+1) = prime(n + 1) * prime(m) * prime(n + 1  m) for some 1 <= m < n + 1. Then, as prime(n + 1) > prime(n) and prime(n + 1  m) > prime(n  m) we have a(n) >= a(n+1) = prime(n + 1) * prime(m) * prime(n + 1  m) > prime(n) * prime(m) * prime(n  m) >= a(n). A contradiction.
%C We contradicted a(n) >= a(n + 1) for n > 1. Therefore for n > 1 we have a(n) < a(n + 1). a(1) = a(2) because prime(0) does not exist.
%C This sequence could help in finding terms for A365280. Once an upper bound is chosen for a search, one could find the largest prime factor that could part of the product prime(m)*prime(t)*prime(m+t) <= u. This way for any prime p > prime(m+t) we do not need to compute primepi(p) saving a bunch of time in checking if a term is in A364462.
%e For n = 2, we take m=1 and get a(2) = prime(1)*prime(1)*prime(2) = 12.
%o (PARI) first(n) = {my(pr = primes(n), res = vector(n, i, oo)); res[1] = 12; for(i = 2, n, for(j = 1, i\2, res[i] = min(res[i], pr[j]*pr[ij])); res[i]*=pr[i]); res} \\ _David A. Corneth_, Aug 31 2023
%Y Cf. A364462, A365280.
%K nonn
%O 1,1
%A _David A. Corneth_, Aug 31 2023
