[seqfan] Questions regarding a new triangle.
Ed Jeffery
lejeffery2 at gmail.com
Wed Nov 21 10:23:36 CET 2012
Hello,
Let d and m be natural numbers such that d <= sqrt(m) and d divides m. Then
there exists a partition of m of the form
m = Sum_{r=0,...,d-1} m/d - d + 1 + 2*r.
The number of such divisors d of m is equal to A038548(m) [5] (and is
always >= 1). More generally, for any natural numbers x,y, there exists a
partition of the product x*y of the form
x*y = Sum_{r=0,...,min(x-1,y-1)} |x - y| + 1 + 2*r,
where |.| means absolute value. For all partitions of this class, the parts
are consecutive odd (respectively, even) when x + y + 1 is odd
(respectively, even). (For the case of either x = 1 or y = 1, we classify a
'singleton' part as 'consecutive.')
Playing around with these partitions while working on some other sequences,
I constructed the following triangle T, with entries
(1) T(n,k) = n! + n - 2*(n - k), 0 <= k <= n:
.....1
.....0.....2
.....0.....2.....4
.....3.....5.....7.....9
....20....22....24....26....28
...115...117...119...121...123...125
...714...716...718...720...722...724...726
..5033..5035..5037..5039..5041..5043..5045..5047
.40312.40314.40316.40318.40320.40322.40324.40326.40328
etc.
This triangle arises from a partition of the product (n+1)*n! = (n+1)! of
the class described above, for n = 0,1,..., namely,
(2) (n+1)*n! = Sum_{k=0,...,min(n!-1,n)} |n! - (n+1)| + 1 + 2*k.
Hence the row sums of T give A000142 [1] up to an offset. For each n, I
originally arranged the terms (parts or summands) from (2) in ascending
order to create rows of a triangle. But the resulting irregular rows for
n=1 and n=2 each just needed to have a zero prepended to make the triangle
a regular one which is why I worked out the version given by (1). There are
a few observations regarding this triangle as follows.
(i) The 'main' diagonal is {T(n,n)} = A005095 [3]. (ii) The first column is
{T(n,0)} = A005096 [4]. (iii) Let S(n) = Sum_{k=0,...,n} (T(n,k))^2, so S =
{1, 2, 20, 164, 2920, 86470, ...}. Then
(3) Lim_{n -> infinity} S(n+1)/S(n) = n*(n+1) = A002378(n) [2],
since obviously S(n) ~ (n+1)*(n!)^2; but is the strict inequality S(n+1)/S(n)
< n*(n+1) always true? Can someone find a generating function for S (or for
T)? Finally, I'd like to know if there is some combinatorial
interpretationof the row entries of T(in terms of enumerating
something useful, e.g., perhaps partitions of some
other classes): does anyone have any ideas regarding this?
Thanks in advance,
Ed Jeffery
References:
[1] OEIS sequence A000142, https://oeis.org/A000142.
[2] OEIS sequence A002378, https://oeis.org/A002378.
[3] OEIS sequence A005095, https://oeis.org/A005095.
[4] OEIS sequence A005096, https://oeis.org/A005096.
[5] OEIS sequence A038548, https://oeis.org/A038548.
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