[seqfan] Re: Proof of conjecture needed.

[Max Alekseyev]

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
>
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