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Template:Sequence of the Day for November 4

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Intended for: November 4, 2011

Timetable

  • First draft entered by M. F. Hasler on November 3, 2011
  • Draft reviewed by Alonso del Arte on April 9, 2012
  • Draft to be approved by October 4, 2012
Yesterday's SOTD * Tomorrow's SOTD

The line below marks the end of the <noinclude> ... </noinclude> section.



The triangle of Eulerian numbers (for concatenated rows
{1, 1, 1, 1, 4, 1, 1, 11, 11, 1, ...}
, see A008292)
n
       
n!
A000142

1   1  
1
2   1 1  
2
3   1 4 1  
6
4   1 11 11 1  
24
5   1 26 66 26 1  
120
6 1 57 302 302 57 1  
720
7   1 120 1191 2416 1191 120 1  
5040

m = 1

2
3
4
5
6
7  

is given by the coefficients of the Eulerian polynomials

En(x)  = 
n
m  = 1
  
E (n, m) xn − m, n ≥ 1,
which appear in the numerator of an expression for the generating function of the sequence
{kn}k  ≥  1 = {1n, 2n, 3n, ...}, n   ≥   1
.
The Eulerian number
E (n, m) =   〈  nm    
is the number of permutations of the numbers 1 to
n
in which exactly
m
elements are greater than the previous element.
The subsequence of Eulerian numbers greater than 1, which are those not lying on the border of the triangle, i.e. with
1 < m < n
, is
{4, 11, 11, 26, 66, 26, ...}
(A014449).

Example

For
n = 4
, the sequence
{kn}n = 4, k  ≥  1
= {1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, ...} = A000583
 (k ), k   ≥   1,
has the generating function
G{k 4, k  ≥  1}(x)  = 
xE4(x)
(1 − x) 5
 = 
x (x 3 + 11 x 2 + 11 x + 1)
1 − 5 x + 10 x 2 − 10 x 3 + 5 x 4x 5
  .