A342678 a(n) is the number of base-2 lunar primes less than or equal to n.

0, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 19, 20, 20, 20, 20, 20, 20, 20, 20, 21, 21, 22, 22, 23, 23, 24

Offset: 1

Author

Ya-Ping Lu, Mar 18 2021

Keywords

• nonn
• base

Comments

a(n) and base-2 lunar prime density, a(n)/n, for some n up to 2^39 are

k n = 2^k a(n) a(n)/n

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

1 2 1 0.5

5 32 11 0.34375

10 1024 323 0.31542...

15 32768 5956 0.35430...

20 1048576 424816 0.40513...

25 33554432 14871345 0.44320...

30 1073741824 502585213 0.46806...

35 34359738368 16593346608 0.48292...

39 549755813888 269325457277 0.48990...

Conjecture: base-2 lunar prime density approaches 0.5 as n tends to infinity, i.e., lim_{n->oo} a(n)/n = 0.5 (see Comments section in A342676).

a(n) is the n-th partial sum of A342704.

Links

• D. Applegate, M. LeBrun and N. J. A. Sloane, Dismal Arithmetic, arXiv:1107.1130 [math.NT], 2011.

Crossrefs

Cf. A342704, A342676, A087097, A087636, A067139, A169912, A171000, A130206, A170806, A171143, A171750, A171752.

Programs

Python

def addn(m1, m2):
    s1, s2 = bin(m1)[2:].zfill(0), bin(m2)[2:].zfill(0)
    len_max = max(len(s1), len(s2))
    return int(''.join(max(i, j) for i, j in zip(s1.rjust(len_max, '0'), s2.rjust(len_max, '0'))))
def muln(m1, m2):
    s1, s2, prod = bin(m1)[2:].zfill(0), bin(m2)[2:].zfill(0), '0'
    for i in range(len(s2)):
        k = s2[-i-1]
        prod = addn(int(str(prod), 2), int(''.join(min(j, k) for j in s1), 2)*2**i)
    return prod
m = 1; m_size = 7; a = 0; L_im = [1]
while m <= 2**m_size:
    for i in range(2, m + 1):
        im_st = str(muln(i, m)); im = int(im_st, 2); im_len = len(im_st)
        if im_len > m_size: break
        if im not in L_im: L_im.append(im)
    if m not in L_im: a += 1
    print(a); m += 1