[seqfan] Re: Is A026477 determined by prime signatures?

[Bob Selcoe]

Hi Charles & Seqfans,

I think that single digit sets (i.e., prime powers p^r) can only
be r = {1,2,4,8,15,22...}, because after p and p^2 appear, r can only 
numbers which cannot be sums of any combination of 1, 2 or 3 smaller members 
of r.

Expanding to prime signatures with
double-digit sets {r,r} (i.e. prime powers (pq)^r), we run into the similar
situation that there can only be certain numbers of this form which appear 
in the
sequence: since p and q appear, pq cannot; since p^2 and q^2 appear, (pq)^2
cannot; since p^4 and q^4 appear and p^8, (pq)^3 must appear but
(pq)^{4,5,6,7,8} cannot, etc.

Unless I'm mistaken, this situation applies to all prime signatures, so I
believe your conjecture is correct.

If so, I'll add it as a comment to A026477.

Cheers,
Bob Selcoe


--------------------------------------------------
From: "Charles Greathouse" <charles.greathouse at case.edu>
Sent: Thursday, August 25, 2016 9:34 AM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Is A026477 determined by prime signatures?

> Sequence A026477 is defined as:
> a(1) = 1, a(2) = 2, a(3) = 3; and for n > 3, a(n) = smallest number >
> a(n-1) and not of the form a(i)*a(j)*a(k) for 1 <= i < j < k < n.
>
> It seems that if two numbers have the same prime signature (multiset of
> prime exponents) then either both or neither are in the sequence, but I
> can't prove this. Anyone? Clark? The version with two numbers (A026422) 
> has
> this property.
>
> It's clear that all primes are in this sequence, which shows that numbers
> of the form pq are not (for distinct primes p and q).
>
> Charles Greathouse
> Case Western Reserve University
>
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> Seqfan Mailing list - http://list.seqfan.eu/
>