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).

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, 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, 9, 16, 18, 17, 2, 19, 1, 10, 10, 11, 3, 18, 19, 2

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.

Links

Table of n, a(n) for n=1..75.

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

Cf. A006530.

Sequence in context: A325828 A200780 A338899 * A194943 A087145 A117172

Adjacent sequences: A362068 A362069 A362070 * A362072 A362073 A362074

Keyword

nonn

Author

Florian Baur, Apr 08 2023

Status

approved