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

[Christopher Gribble]

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