A357378 Lexicographically earliest sequence of positive integers such that the values a(floor(n/2)) * a(n) are all distinct.

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

Offset: 0

Author

Rémy Sigrist, Sep 26 2022

Keywords

• nonn

Comments

See A357379 for the corresponding products.

The sequence is well defined; we can always extend it with a value strictly greater than the square of the greatest value so far.

The sequence is unbounded (otherwise we could only have a finite number of products a(floor(n/2)) * a(n), and therefore a finite number of terms).

For any prime number p: the first term >= p is p.

All prime numbers appear in the sequence.

If a(n) is the first occurrence of some prime number p, then a(m) = 1 for some m in the interval floor(n/2)..2*n.

There are infinitely many 1's in the sequence; as a consequence, every positive integer appears in A357379.

Examples

			The first terms are:
  n   a(n)  a(floor(n/2))  a(floor(n/2))*a(n)
  --  ----  -------------  ------------------
   0     1              1                   1
   1     2              1                   2
   2     2              2                   4
   3     3              2                   6
   4     4              2                   8
   5     5              2                  10
   6     1              3                   3
   7     3              3                   9
   8     3              4                  12
   9     4              4                  16
  10     1              5                   5
  11     3              5                  15
  12     7              1                   7
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..10000
• Rémy Sigrist, Density plot of the first 10000000 terms
• Rémy Sigrist, C program

Crossrefs

Cf. A050292 (additive variant), A357379.

Programs

C

See Links section.

Python

from itertools import count, islice
def agen(): # generator of terms
    alst, disallowed = [1], {1}; yield 1
    for n in count(1):
        ahalf, k = alst[n//2], 1
        while ahalf*k in disallowed: k += 1
        an = k; yield an; alst.append(an); disallowed.add(ahalf*an)
print(list(islice(agen(), 75))) # Michael S. Branicky, Sep 26 2022

Formula

a(2*n + 1) > a(2*n).