[seqfan] Re: A puzzle from Emissary

[Dan Dima]

All this numbers excepting 2^k-1 will contain a 0 and hence eliminate
that 0 you will get a (k-1) length number and we are done.
Not OK


On Mon, Nov 28, 2011 at 9:10 PM, Dan Dima <dimad72 at gmail.com> wrote:
> Proof by induction.
> Suppose this is true by all odd numbers with exactly (k-1) binary digits 0 or 1.
> Show that this is true for all odd numbers with exactly k binary digits.
> All this numbers excepting 2^k-1 will contain a 0 and hence eliminate
> that 0 you will get a (k-1) length number and we are done.
> A separate proof is required for 11..1, k times 2^k-1 but this is simply
> 11...1 + 1 as 2^(k-1)+1 = 2^(k-1)
>
>
>
>
>
> On Mon, Nov 28, 2011 at 8:53 PM,  <franktaw at netscape.net> wrote:
>> A sufficient condition is that every number not already one less than a
>> power of 2 can be transformed into one less than a power of 2 in one step.
>> Is this true? Again, it will suffice to prove it for odd numbers greater
>> than 1.
>>
>> Franklin T. Adams-Watters
>>
>> -----Original Message-----
>> From: Charles Greathouse <charles.greathouse at case.edu>
>> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>> Sent: Mon, Nov 28, 2011 12:38 pm
>> Subject: [seqfan] Re: A puzzle from Emissary
>>
>>
>>> It looks like C=2 and for n>1,
>>> a(n) = 1 if n=2^k a power of 2,
>>> a(n) = 2 otherwise
>>
>> Hmm.  The necessary and sufficient condition needed for that to hold
>> is that any non-power of two can be transformed by the 'binary +
>> insertion' procedure to a power of 2 in one step.
>>
>> In fact, it suffices (and is necessary) to show that this works for
>> odd numbers greater than 1.  Anyone up for a proof by strong
>> induction?
>>
>> Charles Greathouse
>> Analyst/Programmer
>> Case Western Reserve University
>>
>> On Mon, Nov 28, 2011 at 11:52 AM, Dan Dima <dimad72 at gmail.com> wrote:
>>>
>>> It looks like C=2 and for n>1,
>>> a(n) = 1 if n=2^k a power of 2,
>>> a(n) = 2 otherwise
>>>
>>>
>>>
>>> On Mon, Nov 28, 2011 at 6:04 PM, N. J. A. Sloane
>>
>> <njas at research.att.com>
>> wrote:
>>>>
>>>> Dear Seq Fans,
>>>> Victor Miller posted the following on the math-fun list:
>>>>
>>>>> In the lastest issue of Emissary (the newsletter from MSRI) there's
>>
>> the
>>>>>
>>>>> following challenging puzzle:
>>>>>
>>>>> For every positive integer n, write it in binary, and allow the
>>
>> following
>>>>>
>>>>> possible set of transformations:
>>>>>
>>>>> You can insert "+" signs at arbitrary points within the binary
>>
>> expansion,
>>>>>
>>>>> and interpret that as a sum of binary numbers.  For example
>>>>>
>>>>> 110101_2  --> 11_2+01_2+01_2 = 3 + 1 + 1 = 5 = 101_2.
>>>>>
>>>>> Show that there is an absolute constant C such that for any positive
>>>>> integer n there is a sequence of at most C such transformations that
>>>>> results in 1.
>>>>>
>>>>> Note: The minimal value of C is a real shocker
>>>>>
>>>>> This made me wonder about the obvious generalization to other
>>
>> bases, where
>>>>>
>>>>> now we ask for a sequence of such transformation that results in a
>>
>> single
>>>>>
>>>>> digit base b number.
>>>>>
>>>>> Victor
>>>>
>>>> Me: So consider the sequence
>>>> a(n) = smallest number of steps needed to reach 1
>>>>
>>>> I think this begins (for n = 1,2,3,...)
>>>> 0,1,2,1,2,2,2,1,2,2,2,2,2,2,2,1,...
>>>>
>>>> For example, 14 = 1110_2 -> 1 + 11 + 0 = 4 = 100_2 -> 1 + 0 + 0 = 1
>>>> so a(14)=2
>>>>
>>>> Could someone check this and extend it?
>>>>
>>>> Neil
>>>>
>>>>
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