OFFSET
1,7
COMMENTS
The partition method used here is the same as that in A348168.
Conjecture 1: lim_{n->oo} N_i/n = k_i, where N_i is the number of partitions with i elements and k_i is a constant, with k_2 > k_1 > k_3 > k_4 > .... The values of k_i are the same as those in A348168.
Conjecture 2: lim_{n->oo} Sum_{1..n} a(n)/n = lim_{i->oo} Sum_{1..i} i*k_i = e, or the average partition length approaches 2.71828... as n tends to infinity.
Numbers of twin prime pairs (N) and partitions with 1 through 6 twin prime pairs for n up to 10000000 are given in the table below.
n N N_1 N_2 N_3 N_4 N_5 N_6
-------- -------- ------- ------- ------- ------ ------ ------
1 1 1 0 0 0 0 0
10 15 6 3 1 0 0 0
100 209 30 45 16 5 3 1
1000 2536 286 416 145 64 29 19
10000 26474 2851 4331 1271 544 311 190
100000 271338 28034 43375 12923 5731 3002 1870
1000000 2725126 281837 434234 128190 56563 30074 18171
10000000 27120107 2815831 4352926 1276953 563128 302256 181612
EXAMPLE
Twin prime pair averages in the first 10 partitions are: [4], [6], [12], [18], [30], [42], [60, 72], [102, 108], [138, 150], and [180, 192, 198]. Thus, a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = 1, a(7) = a(8) = a(9) = 2, and a(10) = 3.
PROG
(Python)
from sympy import isprime, nextprime; L = [4]
def nexttwin(x):
p1 = nextprime(x); t1 = p1 + 2
while isprime(t1) == 0: p1 = nextprime(t1); t1 = p1 + 2
return p1+1
for _ in range(2, 89):
print(len(L), end = ', ')
t0 = L[-1]; t1 = nexttwin(t0); g0 = t1 - t0; M = [t1]; t = nexttwin(t1); g1 = t - t1
while g1 < g0 and t - t1 <= g1: M.append(t); t1 = t; t = nexttwin(t)
L = M
CROSSREFS
KEYWORD
nonn,new
AUTHOR
Ya-Ping Lu, Feb 15 2025
STATUS
approved