[seqfan] Re: A261009

[Neil Sloane]

Dear Jean-Paul, I added a comment to A261009, based on your message.

(By the way, the line you quote in 3. is part of a Haskell program, I
believe,
which explains why it is hard to interpret!)

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.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com


On Sun, Oct 25, 2015 at 2:48 AM, jean-paul allouche <
jean-paul.allouche at imj-prg.fr> wrote:

> Dear Seqfans
>
> As recalled in http://www.integers-ejcnt.org/p43/p43.mail.html
> [Integers: Powers in prime bases and a problem on central coefficients]
> the following problem, cited in "Concrete Mathematics" by
> Graham, Knuth, and Patashnik, is still open: prove that for all n > 256
> one has that binom(2n,n) is either divisible by 4 or by 9.
>
> This can be easily reduced to: if c_n is the n-th term of A261009,
> then for all k \geq 9, one has
> 2c_k - c_{k+1} \geq 4.
>
> 1. This has been proved up to huge values of k (see ref. above where
>     k = 10^{13} is indicated). This is certainly (as any base change
> problem)
>     a very difficult question... except if some ingenious trick would work:
>     any idea?
>
> 2. This conjecture does not appear in A261009 on the OEIS.
>
> 3. The following appears about this sequence on the OEIS:
>     "a261009 = a053735 . a000079  -- Reinhard Zumkeller, Aug 14 2015"
>     I guess this should read "a261009 = a053735 composed with a000079"
>     as indicated a few lines above:
>     "a(n) = A053735(A000079(n)). - Michel Marcus, Aug 14 2015"
>
> Best wishes
> jean-paul
>
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