[seqfan] Re: A180388-A180389 pops up in unexpected place (Re: PS: Permutation Ascents )

wouter meeussen wouter.meeussen at pandora.be
Tue Nov 9 20:02:15 CET 2010

thanks to R.P. Stanley's slides, I could make a triangular table,
with row sums exactly like A180388, and extending it to n=20.
The conjecture that both are equivalent will need validation by a proof,
which I could never provide.
http://users.telenet.be/Wouter.Meeussen/Table_A180389.txt  (4.3 kb)
Could the original authors check if the triangular table coincides with
the fine-structure of the ascents (count, length,..)?

To get an idea where all this came from, look at
http://users.telenet.be/Wouter.Meeussen/SchurFunctionsDeca.xls (2.5Mb)
(Excel was used for its display facilities, and only minor calculations).
(Most calculation was performed by Mathematica 4.0 linked to Excel)

There are a few conjectures in there :

1    R.P. Stanley's  'hook-content' formula stanley( partition, v ) gives
      coefficient of x^v in the GF for the monomial count in the Schur
      matching that partition as function of the # of variables 'v'

2.    For a given count of variables 'v', the sum of the coefficients of x^v
      all partitions of n follows the GF:  1/( 1-x )^v  /( 1-x^2 )^( v*(
v -1 )/2 )

3.    the sum of the coefficients of x^v for any partition follows a GF of
the form
       Sum(k=1..n ; c_k x^k ) /(1-x)^(n+1)

4.    the sum of the coefficients c_k equals the hook length of the
partition, that is to say,
       it equals the count of Standard Young Tableaux matching that

5.    for a given k, the sum of the coefficients c_k over all partitions of
n follows
       the GF given in A161126 (Triangle read by rows: T(n,k) is the number
       involutions of {1,2,..,n} having k descents (n>=1; 0<=k<n).

6.    and the sum of their squares seems to give A180389 (see title)

Even more conjectures follow from some nice symmetries all over the place.
(hmm, conjectures to me, minor lemma's to others).


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