# [seqfan] Re: Comment on A151750 [erratum]

David Wilson davidwwilson at comcast.net
Sat Aug 1 23:33:10 CEST 2009

```I have to rescind my assertion that

b does not divide choose(2n, n) <=> all b-ary digits of n are < b/2.

for odd b. It is not true for prime b, but not true for 9, and I assume for
any composite. However, I can prove it b = p prime.

Theorem: For prime p

p does not divide choose(2n, n) <=> all p-ary digits of n are < p/2.

Proof: The exponent of prime p in n! is given by

[1] e_p(n) = (n - s_p(n))/(p -1)

where s_p(n) is the sum of the p-ary digits of n (well known to those who
well know it).

So p does not divide choose(2n, n) = (2n)! / (n!)^2 precisely when

e_p(2n) - 2e_p(n) = 0

Together with [1], this implies

s_p(2n) = 2 s_p(n)

But the sum of p-ary digits in 2n is twice the sum of the p-ary digits in n
precisely when there is no carry in the sum n+n, that is to say, every p-ary
digit of n is < p/2, QED.

----- Original Message -----
From: "David Wilson" <davidwwilson at comcast.net>
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Sent: Saturday, August 01, 2009 4:44 PM
Subject: [seqfan] Re: Comment on A151750 [erratum]

Sorry, I got a sign wrong.

----- Original Message -----
From: "David Wilson" <davidwwilson at comcast.net>
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Sent: Saturday, August 01, 2009 4:28 PM
Subject: [seqfan] Re: Comment on A151750

For b >= 2, let

S_b = { n : b does not divide choose(2n, n) }

I believe you can prove

S_b = { n : every digit in b-ary representation of n is <= (b-1)/2 }

This would imply that

|{ n in S_b : n < b^k }| = ((b-1)/2)^k

which would in turn give us a density estimate for S_b.

We can then compute a density estimate S_3 intersect S_5 intersect S_7
intersect S_11 and decide whether we expect this set to be infinite or
finite.

----- Original Message -----
To: <seqfan at list.seqfan.eu>
Sent: Friday, July 31, 2009 3:44 PM
Subject: [seqfan] Comment on A151750

Inspired by Niel's yesterdays submission A151750
I tried to integrate this difficult problem in the
framework of divisibility properties of the swinging
http://www.luschny.de/math/primes/SwingingPrimes.html
---

Let S be a sequence and p a prime number.

S(n) is 'p-primefree' iff no prime <= p divides S(n).

Now we can associate to S

pf(S,i) -> seq(n, such that S(n) is ith_prime-primefree).

For example assume S to be the *odd kernel* of the
swinging factorial \$ (A056040).

Then 6\$ is 3-primefree, 20\$ is 5-primefree, 1512\$
is 7-primefree, 6320\$ is 11-primefree, ...

The first few values of the associated sequences are:

pf(odd(\$),1) = 1,2,6,7,8,18,19,20,24,25,26,54,..
pf(odd(\$),2) = 1,2,20,24,54,60,61,62,72,73,74,504..
pf(odd(\$),3) = 1,2,20,1512,1513,1514,..
pf(odd(\$),4) = 1,2,6320,...
pf(odd(\$),5) = 1,2,...

Which of these sequences are finite? And if they are
finite, what is the largest n such that S(n) is in PF(S,i)?

i=1: infinite (Lucas),
i=2: infinite (Erdös et al.),
i=3: infinite conjectured (?),
i=4: finite conjectured, max=6320,
i>=5: finite? infinite?, max=2 always?
See the comments in A030979 and A151750.

The 3160 from A151750 and of A129489 is 6320 / 2.
A129508, doubled, is a subsequence of pf(odd(\$),2).
A030979, doubled, is a subsequence of pf(odd(\$),3).

A possible attempt to concentrate these relations
in a sequence:

pf(odd(\$)): i -> min {pf(odd(\$),i) > primorial(i)}.
6,20,1512,6320,?,?,..   ( ? = 2 )
======================

Many questions open. Any comment appreciated.

Cheers Peter

--------------------------------------------------
swing := proc(n) option remember;
if n = 0 then 1 elif irem(n, 2) = 1 then
swing(n-1)*n else 4*swing(n-1)/n fi end:

oddprimorial := proc(n) option remember; local i;
mul(ithprime(i+1),i=1..n) end:

pf := proc(n,m) local v,k,p; v := 0;
p := oddprimorial(n);
for k from 1 to m do
if igcd(swing(k),p) = 1 then v := k fi;
if v > p then break fi od; [n,v] end:

submit := proc(n) pf(n,10000) end;
seq(submit(i),i=1..10);

[1, 6], [2, 20], [3, 1512], [4, 6320], [5, 2],
[6, 2], [7, 2], [8, 2], [9, 2], [10, 2], ...
--------------------------------------------------

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