cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A087097 Lunar primes (formerly called dismal primes) (cf. A087062).

Original entry on oeis.org

19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 109, 209, 219, 309, 319, 329, 409, 419, 429, 439, 509, 519, 529, 539, 549, 609, 619, 629, 639, 649, 659, 709, 719, 729, 739, 749, 759, 769, 809, 819, 829, 839, 849, 859, 869, 879, 901, 902, 903, 904, 905, 906, 907, 908, 909, 912, 913, 914, 915, 916, 917, 918, 919, 923, 924, 925, 926, 927, 928, 929, 934, 935, 936, 937, 938, 939, 945, 946, 947, 948, 949, 956, 957, 958, 959, 967, 968, 969, 978, 979, 989
Offset: 1

Views

Author

Marc LeBrun, Oct 20 2003

Keywords

Comments

9 is the multiplicative unit. A number is a lunar prime if it is not a lunar product (see A087062 for definition) r*s where neither r nor s is 9.
All lunar primes must contain a 9, so this is a subsequence of A011539.
Also, numbers k such that the lunar sum of the lunar prime divisors of k is k. - N. J. A. Sloane, Aug 23 2010
We have changed the name from "dismal arithmetic" to "lunar arithmetic" - the old name was too depressing. - N. J. A. Sloane, Aug 06 2014
(Lunar) composite numbers are not necessarily a product of primes. (For example 1 = 1*x for any x in {1, ..., 9} is not a prime but can't be written as the product of primes.) Therefore, to establish primality, it is not sufficient to consider only products of primes; one has to consider possible products of composite numbers as well. - M. F. Hasler, Nov 16 2018

Examples

			8 is not prime since 8 = 8*8. 9 is not prime since it is the multiplicative unit. 10 is not prime since 10 = 10*8. Thus 19 is the smallest prime.
		

Crossrefs

Programs

  • PARI
    A87097=select( is_A087097(n)={my(d); if( n<100, n>88||(n%10==9&&n>9), vecmax(d=digits(n))<9, 0, #d<5, vecmin(d)A087062(m,k)==n&&return))))}, [1..999]) \\ M. F. Hasler, Nov 16 2018

Formula

The set { m in A011539 | 9A054054(m) < min(A000030(m),A010879(m)) } (9ish numbers A011539 with 2 digits or such that the smallest digit is strictly smaller than the first and the last digit) is equal to this sequence up to a(1656) = 10099. The next larger 9ish number 10109 is also in that set but is the lunar square of 109, thus not in this sequence of primes. - M. F. Hasler, Nov 16 2018

A087638 Number of lunar primes with <= n digits.

Original entry on oeis.org

0, 18, 99, 1638, 22095, 264312, 3159111, 36694950, 418286661
Offset: 1

Views

Author

Marc LeBrun and N. J. A. Sloane, Oct 26 2003

Keywords

Comments

Partial sums of A087636. - M. F. Hasler, Nov 15 2018

Crossrefs

Cf. A087062 (lunar product), A087097 (lunar primes), A087636 (#{n-digit primes}).

Programs

Extensions

a(6)-a(9) from David Applegate, Nov 07 2003

A342676 a(n) is the number of lunar primes less than or equal to n.

Original entry on oeis.org

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, 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, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7
Offset: 1

Views

Author

Ya-Ping Lu, Mar 18 2021

Keywords

Comments

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
n 1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000
a(n) 0 0 18 99 1638 22095 264312 3159111 36694950 418286661
a(n)/n 0 0 0.18 0.099 0.164 0.221 0.264 0.316 0.367 0.418
Conjecture 1: Base 10 lunar prime density approaches 0.9 as n tends to infinity, or lim{n->oo} a(n)/n = 0.9.
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.
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.

Crossrefs

Programs

  • Python
    def addn(m1, m2):
        s1, s2 = str(m1), str(m2)
        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 = str(m1), str(m2), '0'
        for i in range(len(s2)):
            k = s2[-i-1]
            prod = addn(int(str(prod)), int(''.join(min(j, k) for j in s1))*10**i)
        return prod
    m = 1; m_size = 2; a = 0; L_im = [9]
    while m <= 10**m_size:
        for i in range(1, m + 1):
            if i == 9: continue
            im_st = str(muln(i, m)); im = int(im_st); 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

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

Original entry on oeis.org

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

Views

Author

Ya-Ping Lu, Mar 18 2021

Keywords

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.

Crossrefs

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

A191366 Number of base 3 lunar primes of length n.

Original entry on oeis.org

0, 4, 4, 20, 59, 164, 544, 1730, 5471, 17765, 55050, 173906, 546893
Offset: 1

Views

Author

David Applegate and N. J. A. Sloane, May 31 2011

Keywords

Crossrefs

Extensions

Additional term from David Applegate, Jun 07 2011

A235641 Number of n-digit lunar primes obtained by promoting the binary templates.

Original entry on oeis.org

0, 18, 81, 1539, 17661, 135489
Offset: 1

Views

Author

Arkadiusz Wesolowski, Apr 20 2014

Keywords

Crossrefs

Showing 1-6 of 6 results.