A364434 a(1) = 12; for n >= 2, a(n) = least positive integer of the form prime(m)*prime(n-m)*prime(n) with m >= 1.

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

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

Table of n, a(n) for n=1..43.

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[i-j])); res[i]*=pr[i]); res} \\ David A. Corneth, Aug 31 2023

Crossrefs

Cf. A364462, A365280.

Sequence in context: A346531 A070710 A048759 * A303646 A298036 A119877

Adjacent sequences: A364431 A364432 A364433 * A364435 A364436 A364437

Keyword

nonn

Author

David A. Corneth, Aug 31 2023

Status

approved