[seqfan] Re: g(k) = (k^2)!*prod(j=0, k-1, j!/(j+k)!)
Meeussen Wouter (bkarnd)
wouter.meeussen at vandemoortele.com
Mon Mar 1 15:50:27 CET 2010
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 n.
I remember this being called 'minor numiracle', big tongue in cheek.
guess you knew this already.
Wouter.
-----Original Message-----
From: seqfan-bounces at list.seqfan.eu [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Christopher Gribble
Sent: maandag 1 maart 2010 0:49
To: 'Sequence Fanatics Discussion list'
Subject: [seqfan] g(k) = (k^2)!*prod(j=0, k-1, j!/(j+k)!)
Dear seqfans,
Please ignore the last email I sent with subject g(k) = (n^2)!*prod(k=0, n-1, k!/(n+k)!) as
it contains several symbolic errors for which I apologise.
In "Mean Values of L-Functions and Symmetry" by J.B. Conrey and D.W. Farmer (1999),
see http://arxiv.org/abs/math/9912107v1, the function g(k) is described as a measure
of "how many polynomials of length T are needed to capture the mean square of
zeta(s)^k" in the Dirichlet polynomial approximation to zeta(s)^k. The first 9 terms in
the sequence for g(k) = (k^2)!*prod(j=0, k-1, j!/(j+k)!) are given in A039622. Conrey
and Farmer indicate that g(k) has an interesting prime factorisation which I have spent
some time investigating and I intend to submit the results to the OEIS.
The Pari code for computing the prime factorisation of g(k) is:
a(k) = factor (if (k<1, 0, (k^2)!*prod(j=0, k-1, j!/(j+k)!)))
The table of exponents in the prime factorisation of g(k), suitably truncated, is
Index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
Prime 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79
k
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
3 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4 3 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5 2 1 1 0 2 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
6 4 2 1 0 2 2 2 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0
7 4 1 1 1 1 2 2 2 2 1 1 1 1 1 1 0 0 0 0 0 0 0
8 7 2 2 1 0 1 3 3 2 2 2 1 1 1 1 1 1 1 0 0 0 0
9 6 4 3 1 0 1 3 4 3 2 2 2 1 1 1 1 1 1 1 1 1 1
10 7 4 4 3 0 0 2 4 4 3 3 2 2 2 2 1 1 1 1 1 1 1
11 6 5 4 3 1 0 2 3 5 4 3 3 2 2 2 2 1 1 1 1 1 1
12 10 7 6 2 1 0 1 2 5 4 4 3 3 3 3 2 2 2 2 2 1 1
13 9 9 6 3 1 1 0 1 4 5 5 4 4 3 3 3 2 2 2 2 2 2
14 11 9 5 4 1 1 0 1 3 6 6 5 4 4 4 3 3 3 2 2 2 2
15 11 8 5 4 2 1 0 0 2 6 7 6 5 5 4 4 3 3 3 3 3 2
16 15 8 5 5 4 1 0 0 2 5 7 6 6 5 5 4 4 4 3 3 3 3
17 14 7 4 6 4 2 1 0 1 4 6 7 7 6 6 5 4 4 4 4 3 3
18 15 8 3 7 3 2 1 0 1 4 5 8 7 7 6 6 5 5 4 4 4 4
19 14 7 3 8 2 4 1 1 0 3 4 8 8 8 7 6 6 5 5 5 4 4
20 17 6 4 9 3 4 1 1 0 2 3 7 9 9 8 7 6 6 5 5 5 5
21 15 8 3 10 3 3 1 1 0 2 3 6 9 10 9 8 7 7 6 6 6 5
22 17 7 3 10 4 3 2 1 0 1 2 6 8 10 10 9 8 7 7 6 6 6
I have discovered (or re-discovered) that if v(p(n), g(k)) denotes the power to which prime(n) is raised in the
prime factorisation of g(k) then it appears that the following partial symmetry holds
v(p(n), g( k)) = v(p(n), g(p(n) - k)), n >= 1, 1 <= k <= (p(n) -
1)/2
Why should this be true ? How can it be proved ?
To see the full extent of the partial symmetry, each factorisation needs to be infinite in extent. Consequently,
I am intending to present the table as a sequence by listing the anti-diagonals read from top to bottom, i.e.
0,0,1,0,0,1,0,0,1,3,0,0,0,1,2,0,0,1,0,1,4,0,0,0,1,1,2,4,0,0,0,1,0,1,1,7,0,0,
0,1,2,0,1,2,6 etc.
All comments are welcome.
Best regards,
Chris Gribble
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