A362017 a(n) is the leading prime in the n-th prime sublist defined in A348168.

2, 3, 5, 7, 11, 17, 23, 29, 37, 53, 59, 67, 79, 89, 97, 127, 137, 149, 157, 163, 173, 179, 191, 197, 211, 223, 239, 251, 293, 307, 331, 347, 353, 359, 367, 397, 409, 419, 431, 439, 449, 457, 479, 521, 541, 557, 587, 631, 673, 683, 691, 701, 719, 787, 809, 821

Offset: 1

Author

Ya-Ping Lu, Apr 04 2023

Keywords

• nonn

Comments

If Conjecture 2 in A348168 is true, lim_{n->infinity} a(n)/prime(round((n-1)*e)+1) = 1, where e is Euler's number.

If a term p (>2) is from a single-prime sublist (A356271), then p is a weak prime (A051635) or a balanced prime (A006562). Otherwise, p is a strong prime (A051634).

The definition divides the primes into maximal sublists such that gaps between adjacent primes in a sublist are smaller than the gap that precedes the sublist and no larger than the first gap within the sublist. - Peter Munn, Jul 07 2025

Examples

			According to the definition in A348168, prime numbers are divided into sublists, L_1, L_2, L_3,..., in which L_n = [p(n,1), p(n,2), ..., p(n,m(n))], where p(n,k) is the k-th prime and m(n) the number of primes in the n-th sublist L_n. Thus, a(n) = p(n,1). The first sublist L_1 = [2]. If p(n,1) <= (prevprime(p(n,1)) + nextprime(p(n,1)))/2, then L_n has only 1 prime, p(n,1). Otherwise, m(n) is the largest integer such that g(n,1) >= g(n,i), where g(n,i) = p(n,i+1) - p(n,i) and 2 <= i <= m(n).
The first 32 primes, for example, are divided into 16 prime sublists:
  [2],
  [3],
  [5],
  [7],
  [11,13],
  [17,19],
  [23],
  [29,31],
  [37,41,43,47],
  [53],
  [59,61],
  [67,71,73],
  [79,83],
  [89],
  [97,101,103,107,109,113],
  [127,131].
The leading primes in these sublists are: 2, 3, 5, 7, 11, 17, 23, 29, 37, 53, 59, 67, 79, 89, 97, 127. Therefore, a(1) = 2, a(2) = 3, ..., and a(16) = 127.
		

Crossrefs

Cf. A006562, A051634, A051635, A348168, A356271.

Programs

Python

from sympy import nextprime; R = [2]; L = [2]
for n in range(2, 57):
    p0 = L[-1]; p1 = nextprime(p0); M = [p1]; g0 = p1-p0; p = nextprime(p1); g1 = p-p1
    while g1 < g0 and p-p1 <= g1: M.append(p); p1 = p; p = nextprime(p)
    L = M; R.append(L[0])
print(*R, sep =', ')