[seqfan] Re: Proof of conjecture needed.

[Neil Sloane]

Max, I've taken the liberty of adding your proof to A257499.

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
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On Fri, May 1, 2015 at 11:56 AM, Max Alekseyev <maxale at gmail.com> wrote:

> 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/
> >>
> >
> >
>
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