[seqfan] Re: g(k) = (k^2)!*prod(j=0, k-1, j!/(j+k)!)

Christopher Gribble chris.eveswell at virgin.net
Mon Mar 1 20:41:09 CET 2010

Hi Wouter,

Thanks very much for reminding me of this alternative way of calculating
the prime exponents of g(k).

Best regards,


-----Original Message-----
From: seqfan-bounces at list.seqfan.eu [mailto:seqfan-bounces at list.seqfan.eu]
On Behalf Of wouter meeussen
Sent: 01 March 2010 5:53 PM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: g(k) = (k^2)!*prod(j=0, k-1, j!/(j+k)!)

hi Chris,

define w[n, b] as (n - sum_of_digits_of_n_in_base_b)/(b-1)
then w[n, p] fro p prime, equals the power of p in the prime decomposition
of  (n!)

since you seek the prime exponents of an expression consisting of a product
of factorials,
you just need to properly sum the prime exponents

 w[n^2, p]+ Sum[k=0...n-1 ; w[k,p]-w[n+k,p] ]

for any n an p. See?


> for some of us who suffer from the Mathematica dialect:
> Table[b=Prime[p];
> w[n^2]+ Sum[w[k]-w[n+k] ,{k,0,n-1}]/.
> w[q_]:>(q-Plus@@IntegerDigits[q,b])/(b-1),{p,22},{n,22}]
> is exactly the table you showed.
> Don't count the trailing zero's of n! in base b, but count the carries in
> I remember this being called 'minor numiracle', big tongue in cheek.
> guess you knew this already.
> Wouter.


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