OFFSET
1,3
COMMENTS
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.
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.
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.
EXAMPLE
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).
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).
PROG
(PARI) gpf(n) = if(n == 1, 1, vecmax(factor(n)[, 1]))
\\ returns the first n terms of the sequence:
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])}
CROSSREFS
KEYWORD
nonn
AUTHOR
Florian Baur, Apr 08 2023
STATUS
approved