A362061 a(1) = 1; for n > 1, a(n) is number of terms in the first n-1 terms of the sequence that have the same number of distinct prime factors as a(n-1).

1, 1, 2, 1, 3, 2, 3, 4, 5, 6, 1, 4, 7, 8, 9, 10, 2, 11, 12, 3, 13, 14, 4, 15, 5, 16, 17, 18, 6, 7, 19, 20, 8, 21, 9, 22, 10, 11, 23, 24, 12, 13, 25, 26, 14, 15, 16, 27, 28, 17, 29, 30, 1, 5, 31, 32, 33, 18, 19, 34, 20, 21, 22, 23, 35, 24, 25, 36, 26, 27, 37, 38, 28, 29, 39, 30, 2, 40, 31, 41, 42

Offset: 1

Author

Scott R. Shannon, Apr 06 2023

Keywords

• nonn

Comments

After 5 million terms the most common numbers for the number of distinct prime factors of the terms are 3, 2, 4, 1, and 5, although it is likely these change as n increases.

See A362062 for the indices where a term with k distinct prime factors first appears.

Examples

			a(9) = 5 as the number of distinct prime factors of a(8) = A001221(a(8)) = A001221(4) = 1, and there are five previous terms, a(3), a(5) a(6), a(7) and a(8), that have one prime factor.
a(11) = 1 as the number of distinct prime factors of a(10) = A001221(a(10)) = A001221(6) = 2, and there is only one term, a(10), that has two prime factors.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20, with a color function representing omega(a(n-1)), where black = 0, red = 1, yellow = 2, ..., magenta = 6.
• Scott R. Shannon, Image of the first 250000 terms.

Crossrefs

Cf. A362062, A362031, A362033, A001221, A354606, A000005, A124056, A342585.

Programs

Mathematica

nn = 120; c[] = 0; j = a[1] = c[0] = 1; m = 0; Do[Set[k, c[m]]; (Set[{a[n], j, m}, {k, k, #}]; c[#]++) &[PrimeNu[k]], {n, 2, nn}]; Array[a, nn] (* _Michael De Vlieger, Apr 06 2023 *)

Python

from itertools import islice
from sympy import primefactors
from collections import Counter
def A362061gen(): # generator of terms
    an, c, d = 1, Counter(), dict()
    while True:
        yield an
        dpf = d[an] if an in d else len(primefactors(an))
        c[dpf] += 1
        an = c[dpf]
print(list(islice(A362061gen(), 81))) # Michael S. Branicky, Apr 06 2023