cp's OEIS Frontend

A362071 a(1) = 1, and thereafter a(n) is the number of terms with index m < n such that gpf(a(m)) = gpf(a(n-1)), where gpf(k) = A006530(k) is the greatest prime factor of k (or 1 if k=1).

%I A362071 #27 Aug 21 2023 22:30:17
%S A362071 1,1,2,1,3,1,4,2,3,2,4,5,1,5,2,6,3,4,7,1,6,5,3,6,7,2,8,9,8,10,4,11,1,
%T A362071 7,3,9,10,5,6,11,2,12,12,13,1,8,13,2,14,4,15,7,5,8,16,17,1,9,14,6,15,
%U A362071 9,16,18,17,2,19,1,10,10,11,3,18,19,2
%N A362071 a(1) = 1, and thereafter a(n) is the number of terms with index m < n such that gpf(a(m)) = gpf(a(n-1)), where gpf(k) = A006530(k) is the greatest prime factor of k (or 1 if k=1).
%C A362071 As the sequence consists of terms that are a count of preceding terms, it is unbounded and its record highs are successive integers. Since a new count begins for every term whose gpf was not in the sequence before, every integer is in the sequence infinitely often.
%C A362071 Choosing a(1) = m != 1 will result in an identical sequence with an offset of 1 until the first occurrence of gpf(m) in the sequence. In the original sequence the next term is 1, whereas in the modified sequence it is 2.
%C A362071 A trivial upper bound is a(n) < n. Is there a tighter bound? The terms are expected to grow with n as the density of primes not yet in the sequence decreases and with it the density of terms equal to 1.
%e A362071 a(3) = 2, because gpf(a(2)) = 1 and there are 2 terms where index m < 3 and gpf(a(m)) = 1, i.e., a(1) and a(2).
%e A362071 a(12) = 5 because gpf(a(11)) = 2 and there are 5 terms where index m < 12 and gpf(a(m)) = 2, i.e., a(3), a(7), a(8), a(10), and a(11).
%o A362071 (PARI) gpf(n) = if(n == 1, 1, vecmax(factor(n)[,1]))
%o A362071 \\ returns the first n terms of the sequence:
%o A362071 A362071UpTon(n) = { my(m = matrix(n,2,a,b,if(b==1,1))); for(i = 2, n, g = gpf(m[i-1,1]); m[i,1] = m[primepi(g)+1,2]++); return(m[,1])}
%Y A362071 Cf. A006530.
%K A362071 nonn
%O A362071 1,3
%A A362071 _Florian Baur_, Apr 08 2023