The case n = 256 is particularly impressive, because (2n choose n) is divisible by neither 3, nor 4, nor 5. This is, by Kummer's theorem, equivalent to the conjunction of: -- n is a power of 2; -- the base-3 expansion of n consists only of digits {0, 1}; -- the base-5 expansion of n consists only of digits {0, 1, 2}. Best wishes, Adam P. Goucher
Sent: Sunday, September 02, 2018 at 3:47 AM From: "Andres Valloud" <avalloud@smalltalk.comcastbiz.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] integers that are either square or thrice a square, where do such beasts hide?
The statement reminded me of Concrete Mathematics' problem 5.112:
5.112: Is \binom{2n}{n} divisible by either 4 or 9, for all n > 4 except n = 64 or n = 256?
This is from 1994 or so. I *think* Lagarias et al solved this circa 1996, and IIRC the answer is yes.
Andres.
On 8/29/18 8:01 , Wouter Meeussen wrote:
do they occur anywhere naturally? in combinatorics or in number theory? I bumped into them by accident looking at bracelet-stuff.
Wouter. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun .
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