A251724 a(1) = 2, and for n>1: a(n) = prime(A251719(n)) * prime(A251719(n) + n - 2), where prime(n) gives the n-th prime.

2, 4, 6, 21, 65, 85, 95, 115, 217, 259, 287, 301, 329, 649, 671, 737, 781, 803, 869, 913, 979, 1067, 1111, 1133, 1177, 1199, 1243, 1703, 1781, 1807, 1937, 1963, 2041, 2119, 2171, 3043, 3077, 3247, 3281, 3349, 3383, 3587, 3791, 3859, 3893, 3961, 4063, 4097, 4267, 4369, 4471, 4573, 4607, 4709, 4777, 4811, 5833, 5909, 5947, 6023, 6289, 6403, 6593, 6631, 6707, 6821, 8579

Offset: 1

Author

Antti Karttunen, Dec 15 2014

Keywords

• nonn

Comments

For n >= 2: a(n) = the first "settled semiprime" in the column n of the sieve of Eratosthenes: a(n) = A083221(A251719(n), n).

The "settling of semiprimes" here means that from that semiprime onward, all the other terms in the same column n of a square array A083221 (which is constructed from the sieve of Eratosthenes) are also semiprimes, obtained by successive iterations of A003961 starting from the semiprime here given as a(n). Cf. comments in A251728 which contains all such semiprimes. The "unsettled" semiprimes are in its complement A138511.

Here we assume that A054272(n), the number of primes in interval [prime(n), prime(n)^2], is nondecreasing (implied for example if Legendre's or Brocard's conjecture is true).

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10351
• Wikipedia, Brocard's Conjecture

Crossrefs

After initial 2, a subsequence of A251728 and A001358.

Cf. A000040, A003961, A054272, A055396, A078898, A083221, A138511, A243055, A251719.

Formula

a(1) = 2; and for n >= 2: a(n) = A000040(A251719(n)) * A000040(A251719(n) + n - 2).

a(n) = A083221(A251719(n), n).

Other identities implied by the definition. For all n >= 1:

A078898(a(n)) = n.

A055396(a(n)) = A251719(n).

For all n >= 2:

A243055(a(n)) = n-2.