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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).

%I A362061 #22 Apr 07 2023 11:27:16
%S A362061 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,
%T A362061 19,20,8,21,9,22,10,11,23,24,12,13,25,26,14,15,16,27,28,17,29,30,1,5,
%U A362061 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
%N 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).
%C A362061 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.
%C A362061 See A362062 for the indices where a term with k distinct prime factors first appears.
%H A362061 Michael De Vlieger, <a href="/A362061/b362061.txt">Table of n, a(n) for n = 1..10000</a>
%H A362061 Michael De Vlieger, <a href="/A362061/a362061_1.png">Log log scatterplot of a(n)</a>, n = 1..2^20, with a color function representing omega(a(n-1)), where black = 0, red = 1, yellow = 2, ..., magenta = 6.
%H A362061 Scott R. Shannon, <a href="/A362061/a362061.png">Image of the first 250000 terms</a>.
%e A362061 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.
%e A362061 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.
%t A362061 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 *)
%o A362061 (Python)
%o A362061 from itertools import islice
%o A362061 from sympy import primefactors
%o A362061 from collections import Counter
%o A362061 def A362061gen(): # generator of terms
%o A362061     an, c, d = 1, Counter(), dict()
%o A362061     while True:
%o A362061         yield an
%o A362061         dpf = d[an] if an in d else len(primefactors(an))
%o A362061         c[dpf] += 1
%o A362061         an = c[dpf]
%o A362061 print(list(islice(A362061gen(), 81))) # _Michael S. Branicky_, Apr 06 2023
%Y A362061 Cf. A362062, A362031, A362033, A001221, A354606, A000005, A124056, A342585.
%K A362061 nonn
%O A362061 1,3
%A A362061 _Scott R. Shannon_, Apr 06 2023