[seqfan] Re: A051404
Vladimir Shevelev
shevelev at bgu.ac.il
Wed Oct 30 19:41:17 CET 2013
Dear Don Reble and SeqFans,
A.-M. Legendre poved that the exponent of the highest power of prime p which divides n! equals (n - s_p(n))/(p-1), where
s_p(n) is the number of digits of n in base p. From this it is easy to obtain that the exponent of the highest power of prime p which divides C(2*n-1, n) is (s_p(n) + s_p(n-1) - s_p(2*n-1))/(p-1). [By the way, if to write this for C(a+b, a), we obtain
s_p(a) + s_p(b) - s_p(a+b) and since it is always>= 0, then for s_p we have a "triangle inequality"]
If p=2 and s_2(n)=k(n) 1-bits, then we have k(n) + k(n-1) - k(2*n-1). But from the equality 2*n-1 = 2*(n-1) +1 we immediately obtain k(n-1) - k(2*n-1) = -1. Thus the exponent of the highest power of 2 which divides C(2*n-1, n) is indeed
equals k(n) - 1. This means that number n is in A051404 if and only if s_2(n) <=2 and s_3(n) + s_3(n-1) - s_3(2*n-1) <= 2.
The second your result is beautiful for computer check. It is interesting that it is directly obtained from a remarkable general result of another classic mathematician E. E. Kummer: the exponent of the highest power of prime p which divides C(N, x) equals to the number of carries which appear in adding x and N-x in base p, or in the considered case - in adding n-1 and n in base 3.
Best regards,
Vladimir
----- Original Message -----
From: Don Reble <djr at nk.ca>
Date: Sunday, October 27, 2013 11:50
Subject: [seqfan] A051404
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Seqfans:
>
> > %I A051404
> > %S
> 1,2,3,4,6,9,10,12,18,33,34,36,40,64,66,192,256,264,272,513,514,516,> 576,768,1026,1056,2304,16392,65664,81920,532480,545259520
> > %N Neither 4 nor 9 divides C(2n-1,n) (almost certainly finite).
> > %A David W. Wilson (davidwwilson(AT)comcast.net)
> > %E Complete up to 2^64 = 18446744073709551616.
>
> What a nice sequence. (Thanks, David.)
>
> Testing whether 4 divides f(n)=C(2n-1,n) is
> surprisingly easy.
> Write n in base 2. If there are k 1-bits, then
> 2^k does not divide
> f(n), but
> 2^{k-1} does divide f(n).
> (You'll enjoy proving this.)
>
> The sequence therefore contains only numbers
> with one or two 1-bits.
>
> The rule for divisibility by 3^k is more
> complicated. Write n in base three. Find each
> maximal substring which does not
> contain a '0' and ends with a '2'. The sum of
> the lengths of those
> substrings equals the exponent of the highest
> power of 3 which
> divides f(n). (You'll _hate_ proving this. Ecch!)
>
> Example: 328979 is "121201021102" in base 3.
> The maximal substrings
> are "1212", "2" (from
> 211), and "2". There are 6 digits in the
> substrings, so 3^6
> divides f(328979), but 3^7 doesn't.
> Example: 68800 is "10111101011" in base 3.
> There are no substrings
> which end in '2', so
> 3 does not divide f(68800).
>
> The sequence therefore contains only numbers
> with no 2-trits,
> or a single 2-trit immediately following a 0-
> trit or at the start.
>
>
> So it's easy to check up to, say, 2^30000. As
> expected, there are
> no more terms in that range.
>
>
> P.S. Just kidding: you'll enjoy finding both
> proofs.
> --
> Don Reble djr at nk.ca
>
>
> --
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Shevelev Vladimir
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