A243358 The densest possibly infinite sequence of primes of the form a(n) = floor[A^(C^n)] for A < 2. The density parameter C here approaches its minimal possible value C_0 = 1.2209864... (A117739), while the corresponding value of A is 1.8252076... (A243370).

2, 2, 2, 3, 5, 7, 11, 19, 37, 83, 223, 739, 3181, 18911, 166657, 2375617, 60916697, 3199316947, 403223394631, 147983594957101, 200280265936061027, 1333721075205083093951, 62146579709944366260614273, 31146685223026045243771057244741

Offset: 1

Author

Andrey V. Kulsha, Jun 03 2014

Keywords

• nonn

Comments

Double-checked by David J. Broadhurst. Terms from a(61) to a(67) from David J. Broadhurst. Terms after a(52) are strong probable primes.

It is very likely, but not yet proved, that the sequence is infinite. However, it is clear that for density parameter C < C_0 = 1.2209864... (see A117739) such a sequence must contain nonprime terms.

Links

• Andrey V. Kulsha, Table of n, a(n) for n = 1..40
• Andrey V. Kulsha and David J. Broadhurst, Table of n, a(n) for n = 1..67
• Chris K. Caldwell, A proof of a generalization of Mills' Theorem

Crossrefs

Cf. A060699, A117739, A243370.

Formula

Once the terms up to the prime 223 are known, the following algorithm works:

1. assign P:=(the largest prime currently in the sequence)

2. assign k:=(the distance between 83 and P in the sequence)

3. assign C:=(logP/log84)^(1/k)

4. assign P:=P^C

5. if floor[P] is prime, add it to the sequence and go to 4

6. add nextprime[P] to the sequence and go to 1

That algorithm gives heuristically as many terms as needed because the increment of C at step 3 becomes so tiny that the values of 84^(C^n) for n < k don't jump over integers anymore (although there's no proof).

So we have a(n) = floor[(84-0)^(C_0^(n-10))], where C_0 = 1.2209864... (see A117739), and "84-0" notation means that when C approaches C_0 from above, the necessary value of A brings A^(C^10) to 84 from below.