

A364434


a(1) = 12; for n >= 2, a(n) = least positive integer of the form prime(m)*prime(nm)*prime(n) with m >= 1.


0



12, 12, 30, 63, 154, 273, 442, 646, 874, 1334, 1798, 2294, 3034, 3526, 4042, 4982, 6254, 7198, 8174, 9514, 10366, 11534, 13114, 14774, 17266, 19594, 20806, 22042, 23326, 24634, 28702, 33274, 35894, 38086, 41422, 44998, 47414, 51182, 54442, 57782, 61934, 64798, 69142
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OFFSET

1,1


COMMENTS

Also a(n) is the least positive integer in A364462 that is divisible by prime(n).
This sequence is strictly increasing for n > 1.
Proof by contradiction:
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.
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.
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.


LINKS



EXAMPLE

For n = 2, we take m=1 and get a(2) = prime(1)*prime(1)*prime(2) = 12.


PROG

(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


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



