# [seqfan] Re: Is 4 a semi-Fibonacci number?

David Wilson davidwwilson at comcast.net
Sat Mar 25 05:09:56 CET 2017

```Also, sF(n) is strictly increasing on odd n, so {sF(2n-1): n >= 1} is the ordered sequence of all range values of sF.  4 is not in that sequence.

> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of
> israel at math.ubc.ca
> Sent: Thursday, March 23, 2017 11:24 AM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: Is 4 a semi-Fibonacci number?
>
> The least n for which any value occurs must be odd, since sF(n) = sF(n/2)
> for even n. So if you have ruled out sF(n) = 4 for odd n, it follows that 4
> can never occur.
>
> Cheers,
> Robert
>
> On Mar 23 2017, Alonso Del Arte wrote:t
>
> >As you know, 4 is not a Fibonacci number. The Fibonacci function can be
> >extended to all real numbers, but to get Fibonacci(x) = 4 requires x be
> >what looks like a transcendental number.
> >
> >But could 4 be a semi-Fibonacci number? (A030067) The definition is sF(1) =
> >1, sF(n) = sF(n/2) if n is even, sF(n) = sF(n - 1) + sF(n - 2) if n is odd.
> >
> >Couple of years ago, Roberg G Wilson v determined that 4 does not occur
> >among the first million terms. It's not a rigorous proof, of course, but it
> >does suggest that 4 never occurs.
> >
> >It's fairly easy to prove that sF(n) = 4 is impossible if n is odd. But I
> >haven't been able to rule out sF(n) = 4 for n even. That would mean sF(n) =
> >8, but the parity of n seems like it could be anything. I can visualize a
> >whole tree but many of the branches of that tree might not even exist.
> >
> >Can anyone make a determination on this question, or is this another one
> of
> >those plausible but unproven conjectures?
> >
> >Al
> >
> >
>
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