[seqfan] Re: More terms are required for A218976
robert.gerbicz at gmail.com
Wed Nov 21 05:53:19 CET 2012
2012/11/20 Remy José Cano Ramírez <aallggoorriitthhmmuuss at gmail.com>
> Would someone be interested in the following problem?...
> The A218976 entry at OEIS has few terms, due mainly to it lacks of a better
> algorithm for generating correct terms than the basic definition of the
> sequence and its testing by trial and error.
> But, I noticed something interesting about this sequence:
> Let be x[i] an element of a table for the first M positive integers. A
> candidate for being accepted as an A218976 term would be tested against
> each one of the M entries at the described table in order to verify if the
> definition of A218976 is satisfied.
> In order to meet this goal, we need the 5-adic valuation of each entry in
> the table. Making it by successive divisions (as ordinarily one might get
> it) consumes too much time. Due to such disadvantage, a replacement
> computation is needed there (if possible).
> My current proposal for replacement is something quite different to the
> Maple program shown there: In some way, a good deal of combinatorial
> structure becomes evident if the successively-repeated outcome frequency is
> counted on the differences between the quantities
> ( floor(x[i]/20) + floor(x[i]/5) )
> valuation(x[i], 5)
> For a wide enough table. My personal goal is at least 1000 terms before Jan
> 1st 2013. But it might be possible only after identified the proper
> algorithm. For now, I am calling what I found "The forty roses gardener
> algorithm", but for not prevent anyone from having the pleasure of watching
> it, I decided omit more specific details here . The first step of an
> algorithmic characterization of the data previously described, is easy to
> catch. The rest of the algorithm is the challenge part where I am sure that
> would need some help. Below is my e-mail address for contact.
> And all the best.
> R. J. Cano, <remy at ula.ve>
> Seqfan Mailing list - http://list.seqfan.eu/
"a(n) is the maximal number such that 10^(2 + floor(k/a(1)) + floor(k/a(2))
+ ... + floor(k/a(n))) divides (k+9)! for all k > 0."
It should be: a(n) is the smallest (positive) integer.
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