[seqfan] Re: Another question

[David Wilson]

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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