A342676 - OEIS

%I #16 Dec 23 2024 03:07:51

%S 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,

%T 2,2,2,2,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,

%U 6,6,6,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,7

%N a(n) is the number of lunar primes less than or equal to n.

%C The density of lunar primes seems to approach a nonzero fraction in contrast to that of the classical primes, which approaches zero as n tends to infinity. a(n) and lunar prime density, a(n)/n, for n up to 10^9 are

%C   n       1 10  100  1000 10000 100000 1000000 10000000 100000000 1000000000

%C   a(n)    0  0   18    99  1638  22095  264312  3159111  36694950  418286661

%C   a(n)/n  0  0 0.18 0.099 0.164  0.221   0.264    0.316     0.367      0.418

%C Conjecture 1: Base 10 lunar prime density approaches 0.9 as n tends to infinity, or lim{n->oo} a(n)/n = 0.9.

%C D. Applegate, M. LeBrun and N. J. A. Sloane conjectured that the number of base b lunar primes with k digits approaches (b-1)^2*b^(k-2) as k tends to infinity. And necessary conditions for a number n to be prime are that it contain b-1 as a digit and (if k > 2) does not end with 0 (see Links). Since the number of base b integers with k digits equals b^k - b^(k-1), the lunar prime density among integers with k digits should be (b-1)^2*b^(k-2)/(b^k - b^(k-1)), which is 1 - 1/b as k -> oo, if the conjecture holds. Note that, as b increases, the limit approaches 1, or lim_{b->oo} lim_{n->oo} a(n)/n = 1. As n tends to infinity, the probability of finding a base b number having a digit of b-1 approaches 100%, and the probability of finding a base b number ending with 0 approaches 1/b. Therefore, essentially all numbers except those ending with 0 are lunar primes as n tends to infinity.

%C Conjecture 2: Base b lunar prime density approaches 1 - 1/b as n tends to infinity, or lim{n->oo} a(n)/n = 1 - 1/b.

%H D. Applegate, M. LeBrun and N. J. A. Sloane, <a href="http://arxiv.org/abs/1107.1130">Dismal Arithmetic</a>, arXiv:1107.1130 [math.NT], 2011.

%o (Python)

%o def addn(m1, m2):

%o     s1, s2 = str(m1), str(m2)

%o     len_max = max(len(s1), len(s2))

%o     return int(''.join(max(i, j) for i, j in zip(s1.rjust(len_max, '0'), s2.rjust(len_max, '0'))))

%o def muln(m1, m2):

%o     s1, s2, prod = str(m1), str(m2), '0'

%o     for i in range(len(s2)):

%o         k = s2[-i-1]

%o         prod = addn(int(str(prod)), int(''.join(min(j, k) for j in s1))*10**i)

%o     return prod

%o m = 1; m_size = 2; a = 0; L_im = [9]

%o while m <= 10**m_size:

%o     for i in range(1, m + 1):

%o         if i == 9: continue

%o         im_st = str(muln(i, m)); im = int(im_st); im_len = len(im_st)

%o         if im_len > m_size: break

%o         if im not in L_im: L_im.append(im)

%o     if m not in L_im: a += 1

%o     print(a); m += 1

%Y Cf. A087097, A087636, A067139, A087638, A169912, A171000, A130206, A170806, A171143, A171750, A171752.

%K nonn,base

%O 1,29

%A _Ya-Ping Lu_, Mar 18 2021