A071810

[Dan Dima]

You wrote:

2) Can anybody show that a(n+1) < c*a(n) for some c < 2 and n
sufficiently large?
3) Note that (2) suffices to prove the limit above.  Failing that, some
other proof of the limit would be nice.


Note that (2) will never proof the limit above.
This one suffices: a(n+1) > c*a(n) for some c > 1 and n sufficiently large.



On 9/8/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
> This sequence (Number of subsets of the first n primes whose sum is a
> prime) has the comment "... a(n+1) < 2*a(n). Therefore Lim -> oo,
> a(n)/2^n = 0".  I have two problems with this:
>
> 1) It is not obvious that a(n+1) < 2*a(n).
> 2) The "therefore" does not follow.
>
> So,
>
> 1) Can anybody show that a(n+1) < 2*a(n) [for n > 1].
> 2) Can anybody show that a(n+1) < c*a(n) for some c < 2 and n
> sufficiently large?
> 3) Note that (2) suffices to prove the limit above.  Failing that, some
> other proof of the limit would be nice.
> 4) Can anybody evaluate lim_{n->infinity} a(n+1)/a(n)?
>
> Franklin T. Adams-Watters
>
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