Apologies, then, for being a day late. At any rate, I was curious as to the growth rate of (odd indexed elements) of sF, so I computed a few values of sF(2^k - 1). Looks to grow pretty fast, perhaps like n^(c log(n)). n sF(n) 1 1 3 2 7 5 15 16 31 69 63 430 127 4137 255 64436 511 1676353 1023 74555322 2047 5777029421 4095 792086153688 8191 194591768192733 16383 86534148901444102 32767 70244955881077121873 65535 104827174339054175240700 131071 289320796542222620694103961 262143 1484554051525798547483095566354 524287 14226732533670039175251226319995157 1048575 255658362329675830634139832019198358176 2097151 8646156383344527622677186248446917786683381 4194303 552073085216495053730117030388698424955408298334 8388607 66749153256722612438012197449644764555811889274235129 16777215 15322089889433495509821892166057427813245442064758958688836 33554431 6693576684186674844436583641024041055004958124882988723321728369 67108863 5577272394534357747969836555204334929788023962243060637883527988942186 134217727 8881483115234363964483793627146185513254764624755802869033890445863234590077 268435455 27080472748266980783930080876503920295698702610405257020769093391323735487516698472 536870911 158372376484792407948495752863706348514435815566310070102912509445802547425058287520342413 From: maximilian@hasler.fr [mailto:maximilian@hasler.fr] On Behalf Of M. F. Hasler Sent: Saturday, March 25, 2017 5:44 PM To: David Wilson Subject: Re: [seqfan] Re: Is 4 a semi-Fibonacci number? On Sat, Mar 25, 2017 at 12:09 AM, David Wilson <davidwwilson@comcast.net> wrote: 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. That's what I said a day ago... <http://list.seqfan.eu/pipermail/seqfan/2017-March/017390.html> ;-) M.
-----Original Message----- From: SeqFan [mailto:seqfan-bounces@list.seqfan.eu] On Behalf Of israel@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