A348168 Segment the list of prime numbers into sublists L_1, L_2, ... with L_1 = {2} and L_n = {p_1, p_2, ..., p_a(n)}, where a(n) is the largest m such that for 0 < i < m, p_1 - prevprime(p_1) > p_2 - p_1 >= p_{i+1} - p_i.

1, 1, 1, 1, 2, 2, 1, 2, 4, 1, 2, 3, 2, 1, 6, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 4, 2, 8, 1, 4, 2, 2, 1, 1, 5, 2, 1, 2, 2, 2, 1, 4, 6, 2, 2, 5, 8, 7, 2, 1, 1, 2, 10, 2, 2, 2, 2, 1, 4, 4, 2, 1, 5, 2, 1, 1, 2, 2, 2, 2, 1, 2, 2, 1, 4, 1, 1, 3, 2, 2, 3, 1, 2, 1, 2, 1, 2

Offset: 1

Author

Ya-Ping Lu, Oct 03 2021

Keywords

• nonn
• easy

Comments

The gap between two consecutive primes in L_n is smaller than g_{n-1} and g_n, where g_n is the gap between L_n and L_{n+1}. Sublists of length 2 are the most frequent ones and any pair of twin primes >= 11 stay in the same sublist.

Conjecture 1: lim_{n->oo} N_i/n = k_i, where N_i is the number of the first n sublists consisting of i primes and k_i is a constant, with k_2 > k_1 > k_3 > k_4 > ... .

Conjecture 2: lim_{n->oo} (Sum_{i=1..n} a(i))/n = Sum_{i=1..oo} i*k_i = e, meaning that, as n tends to infinity, the average length of sublists approaches 2.71828... (see the partial average - n plot in the links).

From Ya-Ping Lu, Apr 15 2024: (Start)

The distribution of sublists with 1, 2, 3, 4 and 5 primes and the number of primes in the first n sublists are given in the table below. k_i's as defined in Conjecture 1 are: k1 = 0.281, k2 = 0.431, k3 = 0.127, k4 = 0.058, and k5 = 0.031, approximately. Sublists with length <= 5 account for about 93% of the terms and 70% of the primes, as n approaches infinity.

n N_1 N_2 N_3 N_4 N_5 # of primes

---------- --------- --------- --------- -------- -------- -----------

1 1 0 0 0 0 1

10 6 3 0 1 0 16

100 33 44 5 9 3 232

1000 277 431 120 72 36 2617

10000 2821 4225 1243 642 331 27214

100000 28072 42929 12427 6059 3159 276081

1000000 279751 430299 126008 59729 32043 2747392

10000000 2804959 4303512 1264532 592726 317127 27426366

100000000 28070302 43078975 12686566 5869443 3143266 273972452

1000000000 280903920 431182582 127100032 58293618 31258182 2737643048

(End)

Examples

			See also the table of the sublists in the examples for A362017.
a(1) = 1 because L_1 = {2} by definition.
In the following examples we use p_0 to denote prevprime(p_1).
a(2) = 1. For the 2nd sublist, p_1 - p_0 = 3 - 2 = 1. If the next prime, 5, is in L_2, then p_2 - p_1 = 2 > p_1 - p_0. Therefore, 5 does not belong to L_2 and L_2 = {3}.
a(5) = 2. For the 5th sublist, p_1 - p_0 = 11 - 7 = 4. p_2 = 13 is in L_5 because p_2 - p_1 = 2 < p_1 - p_0. However, the next prime, 17, is not in L_5 as 17 - 13 > p_2 - p_1. Thus, L_5 = {11, 13}.
a(15) = 6. L_15 = {97, 101, 103, 107, 109, 113}, because p_1 - p_0 = 97-89 > p_2 - p_1 = 101-97 = 4, which is the maximum prime gap in L_15. 127, the prime after 113, is not in L_15 as 127-113 = 14 > p_2 - p_1.
		

Links

• Ya-Ping Lu, Partial average of A348168
• Index entries for sequences related to gaps between primes

Crossrefs

Cf. A362017 (first in each sublist), A087641, A226657, A001359, A023200.

Programs

Python

from sympy import nextprime
L = [2]
for n in range(1, 100):
    print(len(L), end =', ')
    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

Extensions

Edited by Peter Munn, Jul 08 2025