A355269 - OEIS

A355269

Lexicographically earliest infinite sequence of distinct positive integers such that a(n+1) is prime to the number of divisors of a(n).

1, 2, 3, 5, 7, 9, 4, 8, 11, 13, 15, 17, 19, 21, 23, 25, 10, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 14, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 6, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119

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OFFSET

1,2

COMMENTS

1,2,3 are the earliest consecutive numbers satisfying the definition, therefore these are the initial terms. The sequence is infinite since there is always a number prime to d(a(n)), and we take the least such as a(n+1).

Tau(k) is odd iff k is a square, therefore if a(n) is nonsquare, a(n+1) is odd. Consequently the sequence displays runs of consecutive odd nonsquare numbers, extending between successive odd squares, whereupon the parity of tau(a(n)) changes from even to odd. While odd numbers occur early, even numbers are very delayed because (in general) an even number can only be admitted following an odd square, or (on rarer occasions) following an even square subsequent to an odd square (the only time we observe two consecutive even terms).

Conjectures: When a(n) is square a(n+1) is even; no two squares of equal parity can appear as adjacent terms; sequence is a permutation of the positive integers with primes in natural order.

From Michael De Vlieger, Jun 28 2022: (Start)

Let j = a(n-1), k = a(n), tau(m) = A000005(m), and P(m) = A002110(m).

Theorem: prime p | tau(j) implies gcd(p, k) = 1. Consequence of sequence definition.

Even tau(j) implies nonsquare j (in A000037), which in turn implies odd k. Prime j implies tau(j) = 2, which in turn implies odd k. Therefore there is a significant tendency for k to be odd.

Square j (in A000290) implies odd tau(j) but do not necessarily lead to even k. For n > 3, the smallest missing numbers are even, therefore even k is likely to enter the sequence following square j.

Terms j such that tau(j) is coprime to 6, i.e., j in A352475, allow successor k such that 6 | k. For n > 7, the smallest missing number is divisible by 6 because we usually have gcd(tau(j), 6) > 1. Therefore k such that 6 | k likely follows j such that tau(j) is coprime to 6. A similar argument can be made regarding j such that tau(j) is coprime to 30, i.e., j in A354178.

Primorials appear latest in this sequence because they are divisible by the smallest primes. For example, a(1132) = 30 and a(4884401) = 210.

Generally, we have k : k | P(m) iff j : (tau(j), P(m)) = 1. Since gcd(tau(j), P(m)) > 1 is more common than gcd(tau(j), P(m)) = 1 for m > 1, we are apt to see k | P(m). (End)

LINKS

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Michael De Vlieger, Annotated log log scatterplot of a(n), n = 1..2^16, showing records in red and local minima in blue. Three prominent trajectories appear; that of odd and early terms mostly in red, that of terms divisible by 6 mostly in blue, and even numbers that are not divisible by 6 in green. (It may be that numbers divisible by 6 are not always minima; instead, numbers divisible by a larger primorial may be delayed to a greater degree than those merely divisible by 6 for large n.)

EXAMPLE

a(4)=5 because a(3)=3 has 2 divisors and 5 is the least unused term prime to 2. Likewise a(5) = 7, and a(6) = 9, the first odd square. Since 9 has 3 divisors a(7) is 4, the smallest unused number prime to 3, and the first occurrence of an even square, also having 3 divisors. Consequently a(8)=8, the smallest unused number prime to 3. Thus we see 4,8 the first occasion of a pair of adjacent even terms, consequence of odd square (9) followed by even square (4). The next occasion is a(67,68,69)=121,16,12.

MATHEMATICA

nn = 2^16; a[1] = c[1] = j = 1; u = 2; v = 3; Do[k = u; If[EvenQ[j], k = v; While[Nand[c[k] == 0, CoprimeQ[j, k]], k += 2], While[Nand[c[k] == 0, CoprimeQ[j, k]], k++]]; Set[{a[i], c[k]}, {k, i}]; j = DivisorSigma[0, k]; If[k == u, While[c[u] != 0, u++]]; If[k == v, While[c[v] != 0, v += 2]], {i, 2, nn}]; Array[a, nn] (* Michael De Vlieger, Jun 28 2022 *)

PROG

(Python)

from math import gcd

from sympy import divisor_count

from itertools import count, islice

def agen(): # generator of terms

aset, k, mink = {1, 2}, 1, 3; yield from [1, 2]

for n in count(3):

an, k = k, mink

while k in aset or not gcd(divisor_count(an), k) == 1: k += 1

aset.add(k); yield k

while mink in aset: mink += 1