[seqfan] Re: Proof of conjecture needed.

[Max Alekseyev]

Ed has spotted a couple of typos in my proof. So, just for the record, here
is a corrected version:

Proof. Reformulating the conjecture, one needs to prove that for any
integer m>=0, the equation
(1 + 2^n*(6*k - 3 + 2*(-1)^n))/3 = 2*m + 1
has a unique solution in integers n,k >= 1.

Simplifying a bit, we have:
2^n*(6*k - 3 + 2*(-1)^n) = 6*m + 2.
Since in parentheses we have an odd number, the value of n is uniquely
defined as n = A007814(6*m+2). Since 6*m+2 is even, we have n>=1.

Dividing by 2^n and shuffling the terms, we further get:
6*k = (6*m + 2)/2^n + 3 - 2*(-1)^n.
It remains to prove that the right hand side is divisible by 6.
First, the value of n implies that (6*m + 2)/2^n is odd, and hence the
r.h.s. is even and thus divisible by 2.

Second, taking the r.h.s. modulo 3 we get
(6*m + 2)/2^n + 3 - 2*(-1)^n == 2/(-1)^n + 0 - 2*(-1)^n == 0  (mod 3).
That is, the r.h.s. is divisible by both 2 and 3, implying that integer k
is uniquely defined as:
k = ( (6*m + 2)/2^n + 3 - 2*(-1)^n ) / 6.
It is easy to see that (6*m + 2)/2^n >= 1, and thus k >= 1.

Q.E.D.

Regards,
Max


On Wed, Apr 29, 2015 at 5:21 PM, Max Alekseyev <maxale at gmail.com> wrote:

> Hi Ed,
>
> Reformulating your conjecture, one needs to prove that for any integer
> m>=0, the equation
> (1 + 2^n*(6*k - 3 + 2*(-1)^n))/3 = 2*m + 1
> has a unique solution in integers n,k >= 1.
>
> Simplifying a bit, we have:
> 2^n*(6*k - 3 + 2*(-1)^n) = 6*m + 2.
> Since in parentheses we have an odd number of the value of n is uniquely
> defined as n = A007814 <https://oeis.org/A007814>(6*m+2). Since 6*m+2 is
> even, we have n>=1.
>
> Dividing by 2^n and shuffling the terms, we further get:
> 6*k = (6*m + 2)/2^n + 3 - 2*(-1)^n.
> It remains to prove that the right hand side is divisible by 6. First, the
> value of n implies that (6*m + 2)/2^n is odd, and hence the r.h.s. is even
> and thus divisible by 3.
> Second, taking the r.h.s. modulo 3 we get
> (6*m + 2)/2^n + 3 - 2*(-1)^n == 2/(-1)^n - 2*(-1)^n == 0  (mod 3).
> That is, the r.h.s. is divisible by both 2 and 3, implying that integer k
> is uniquely defined as:
> k = ( (6*m + 2)/2^n + 3 - 2*(-1)^n ) / 3.
> It is easy to see that (6*m + 2)/2^n >= 1, and thus k >= 1.
>
> Q.E.D.
>
> Regards,
> Max
>
>
>
>
>
>
>
>
> On Wed, Apr 29, 2015 at 12:05 PM, L. Edson Jeffery <lejeffery2 at gmail.com>
> wrote:
>
>> Consider the rectangular array A (https://oeis.org/draft/A257499)
>> beginning
>>
>>  1     5     9    13    17
>>  7    15    23    31    39
>>  3    19    35    51    67
>> 27    59    91   123   155
>> 11    75   139   203   267
>>
>> A007583 and A136412 (omitting the initial 2) are bisections of of the
>> first
>> column. The entry in row n and column k of A is given by
>>
>>   A(n,k) = (1 + 2^n*(6*k - 3 + 2*(-1)^n))/3  (n,k >= 1).
>>
>> I have a result that depends on the following
>>
>> Conjecture: The rows (or columns) of A are pairwise disjoint, and their
>> union exhausts the odd natural numbers without duplication; or,
>> equivalently, the sequence A257499 (draft) is a permutation of the odd
>> natural numbers.
>>
>> Would someone like to prove this conjecture and relay the result to me
>> (either here or privately)?
>>
>> Ed Jeffery
>>
>> _______________________________________________
>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>