[seqfan] Re: Another question
Maximilian Hasler
maximilian.hasler at gmail.com
Wed Nov 5 23:56:59 CET 2008
I also think
%N A029909 Starting with n (but omitting the primes), repeatedly sum
prime factors (counted with multiplicity) until reaching a limit.
is not very well posed. On one hand, it should better say:
"starting with A018252(n) ..."
and it should say what is meant by "reaching a limit"
(I suppose: a fixed point... ok, in some sense this is "a" limit - in
fact, *the* limit of the sequence, but well....).
As to your question:
> %C A029909 Is this sequence generating ALL prime numbers greater than 3?
The answer is yes, without assuming Goldbach's conjecture.
I fact even the *weak* Goldbach conjecture,
http://en.wikipedia.org/wiki/Goldbach%27s_weak_conjecture
would be sufficient, but clearly the requirement is even much weaker :
since an arbitrary number of primes can be used, we can simply
subtract from N, if it exceeds the range (~10^18?) up to which the
weak G.conjecture is verified, a sufficiently large multiple K of any
other prime P, and then take the product of the first 3 primes and
P^K, to get an antecedant of N for the considered iteration.
Maximilian
On Wed, Nov 5, 2008 at 2:29 PM, David Wilson <dwilson at gambitcomm.com> wrote:
> Let sop(n) be the sum of primes of n with multiplicity.
>
> Assuming Goldbach is true, every even number n >= 4 is of the form q+r
> where q and r are prime.
> Then every prime >= 7 is of the form p = 3+q+r. We then have p =
> A029909(3qr), so every prime >= 7
> is in A029909. Also A029909(6) = 5, so every prime >= 5 is in A029909.
>
> Let n >= 5 and let A029909(n) = p. Again assuming Goldbach, n is of the
> form q+r or q+r+s with
> q, r, s prime. We can also show that qr or qrs > n. This means that
> A029909(qr or qrs)
> = A029909(q+r or q+r+s) = A029909(n). So if A029909(n) = p, there is a
> larger number j
> (namely qr or qrs) with A029909(n) = p. Thus there are an infinite
> number of n with A029909(n) = p.
>
> Note that by the definition, we should have A029909(1) = A029909(sum of
> prime factors of 1)
> = A029909(0) (since 1 has the empty prime factorization, whose sum is 0)
> = undefined (since 0
> does not have a prime factorization). Hence 1 should not technically be
> in the domain of this sequence,
> unless you legislate A029909(1) = 0.
>
> As it stands, the range of A029909 is {0, 4} U PRIMES >= 5 (though I
> disagree with 0 appearing
> in the sequence). 0 and 4 each occur just once, the primes all occur an
> infinitude of times (all modulo
> the Goldbach conjecture).
>
> Alexander Povolotsky wrote:
>> Here is another question (actually two questions ...),
>>
>> %C A029909 Is this sequence generating ALL prime numbers (greater than 3)
>> ? Also how many times each prime (greater than 3) is generated in this
>> sequence?
>> [From Alexander R. Povolotsky (pevnev(AT)juno.com), Nov 05 2008]
>>
>> Regards,
>> ARP
>>
>>
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