[seqfan] Re: Is 4 a semi-Fibonacci number?
M. F. Hasler
oeis at hasler.fr
Fri Mar 24 15:14:17 CET 2017
On Thu, Mar 23, 2017 at 11:23 AM, <israel at math.ubc.ca> wrote:
> 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.
This more or less settles the issue, but one could add that the odd indexed
terms are by their mere definition (a(2n+1)=a(2n)-a(2n-1)=a(2n-1)+a(n))
strictly increasing. Thus, that bissection (A030068) is also the range of
For completeness I tentatively submitted oeis.org/draft/A284282
<https://oeis.org/A284282> which yields the k such that sF(2k-1) = n, or 0
if n does not occur.
Setting nonzero values to 1, this is also a characteristic function of
A030068 (up to the offset).
> 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
>> 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
>> 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
>> those plausible but unproven conjectures?
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