[seqfan] Re: Several remarks on A088855

[Vladimir Shevelev]

Dear Seq Fans,

I suggest to name the sequence
A010551 also "Peretz factorial" after
Avi Peretz.
In 2001, in  [formula in A010551] 
it seems he for the first time
noted that A010551 is the number of
permutations p of {1, 2, 3, ..., n} 
such that for every i, i and p(i) have
the same parity, i.e., p(i) - i is even. 
Although he did not give a proof,
it is easily reached by induction.
Using Peretz factorial, I obtained
the property 2) of entries of the
triangle A088855.

Unfortunately, I did not find any
his paper. Maybe, anyone
knows?

Best regards,
Vladimir

________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 19 November 2017 21:17
To: seqfan at list.seqfan.eu
Subject: [seqfan] Several remarks on A088855

Dear Seq Fans,

A088855 is a fine triangle.

I would like to show its remarkable
properties which were not noted in A088855
:

1) It is convenient for me to make the
numbering of rows and columns as in
Pascal's triangle,
that is, starting with 0.
So we change
n by n+1, k by k+1 (although
we retain the notation a(n,k) for the
entries).

Then the formula in A088855
for entries a(n,k) of the triangle
could be converted to an interesting
form. Let us denote by n!^ the number
A010551(n). Then
a(n,k) = n!^/(k!^*(n-k)!^).

 This expression we denote
naturally by (binomial^)(n,k).
In this form a(n,k) looks
like as an analog of the number of
k-combinations.
 And this is justified
according to the following
reason
:

2) A combinatorial sense
:
a(n,k) is the number of those
k-combinations when we
choose  k elements {n_1,..., n_k}
from 1..n for which n_1 is odd,
n_2 is even,..., n_k has the
parity of k. For example,
a(6,3) = 9, since we have
only the following suitable
triples
:
{1,2,3},{1,4,3},{1,6,3},
{1,2,5},{1,4,5},{1,6,5},
{3,2,5},{3,4,5},{3,6,5}.

3) Most of entries a(n,k)
are obtained by Pascal rule
:
a(n,k) = a(n-1, k-1) + a(n-1,k),
except for the cases when
n is even and k is odd, but even
in the exceptional cases we
have an "almost Pascal rule"
:
a(n,k) =(n/(n+2))*
(a(n-1, k-1) + a(n-1,k)), n>=2.

4) A generalization.

Note that there are infinitely many
triangles with close properties.
If instead factorial to consider
sequence A_t : multiply
successively by 1,...,1,2,...,2,3,...,3,4, ...,4,...,
n >= 1, a(0) = 1, where we have
t times of every i, i>=1, then the case
t=1 corresponds to Pascal's triangle,
t=2 corresponds to A088855,
t=3
corresponds to the following triangle

1
1 . 1
1 . 1 . 1
1 . 1 . 1 . 1
1 . 2 . 2  . 2 . 1
1 . 2 . 4  . 4 . 2 . 1
1 . 2 . 4  . 8 . 4 . 2 . 1
1 . 3 . 6 . 12 .12 . 6 . 3 . 1
....

such that in OEIS there are neither triangle
nor its row sums, etc.


Best regards,
Vladimir

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