[seqfan] Re: Proof of conjecture needed.
Henry Ricardo
odedude at yahoo.com
Thu Apr 30 02:34:18 CEST 2015
The conjecture is true. My proof is different from Max's. I show that (1) the function A(n, k) is onto and (2) the function is one-to-one.The 'onto' proof is constructive:
Let x be any positive odd integer and set x = A(n, k). Then 3x - 1 = (2 ^ n)(6k - 3 +2 (-1) ^ n), an even integer.Let 2^ r be the largest power of 2 that's a factor of 3x - 1. Then 3 x - 1 = (2 ^ r) y = (2 ^ n)(6k - 3 +2 (-1) ^ n), where y is odd.
Thus y = 2 ^ (n - r) (6k - 3 + 2 (-1) ^ n), which is impossible unless n - r = 0, or n = r.Now we have y = (6k - 3 + 2 (-1) ^ n), or k = (y + 3 - 2(-1) ^ n). Thus the odd integer x appears in row r , column k of the array.
Henry Ricardo
New York Math Circle
From: Max Alekseyev <maxale at gmail.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Wednesday, April 29, 2015 5:21 PM
Subject: [seqfan] Re: Proof of conjecture needed.
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/
>
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