[seqfan] Re: finite a(n) for q=0 and infinite a(n) for q>0?

[Allan Wechsler]

I confess that I am confused. I thought I understood Yuri's definition, and
my painful hand calculations agree with the first few values for q=1. But I
don't see why q=0 should produce a finite  sequence.

Each q selects a set of "good" divisors for every k. The "good" divisors
are the ones that are the arithmetic mean of two primes, separated by 2q.
If q = 0, this requirement reduces to the requirement that the divisor
itself is prime. So for q=0, a(n) should be the smallest integer with n
prime divisors. This sequence is easy to compute: it's the famous
"primorial" sequence A002110, and it is obviously infinite.

I conclude that I have not correctly understood Yuri's definition -- so why
am I getting the same numbers for q=1?

On Mon, Oct 22, 2018 at 1:30 PM юрий герасимов <2stepan at rambler.ru> wrote:

> Dear seqfans, if a(n) is the smallest k such that number of divisors m of
> k is
> equal to n where m-q and m+q are both prime, then a(n) is finite sequence
> for q=0
> and a(n) is infinite sequence for q>0? Example: a(n) = 2 for q=0; a(n) =
> 4, 18,
> 12, 192, 60, 240, 180, 2088, 3360, 4800, 1620, 6300, 7560, ... for q=1;
> a(n) = 5,
> 15, 165, 45, 225, 405, 315, 1575, 2205, 5985, 3465, 10395, ... for q=2,
> e.t.c.Thanjs.
> JSG.
>
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