OFFSET
0,5
COMMENTS
For the symmetric functions a_k see a comment in A196841.
The definition of the family of number triangles
S_{i,j}(n,k),n>=k>=0, 1<=i<j<=n+2, has been given in
A196845. The present triangle is S_{3,4}(n,k) (no 3 and 4
admitted). The first three lines coincide with those of
triangle A094638(n+1,k+1) which tabulates a_k(1,2,...,n).
FORMULA
a(n,k) = 0 if n<k, a(0,0) = 1, a(1,k) = a_k(1) for k=0,1, a(2,k) = a_k(1,2) for k=0,1,2, and a(n,k) = a_k(1,2,5,6,...,n+2), n>=3; k=0..n, with the elementary symmetric functions a_k (see the comment above).
a(n,k) = |s(n+1,n+1-k)| for 0<=n<3,
a(n,k) = sum(((3*4)^m)*(|s(n+3,n+3-k+2*m)| - (3*S_3(n+1,k-1-2*m) + 4*S_4(n+1,k-1-2*m))),m = 0..floor(k/2)), with the Stirling numbers of the first kind s(n,m) = A048994(n,m), and the number triangles S_3(n,k)= A196842(n,k) and S_4(n,k)= A196843(n,k) (for negative k one puts the entries of these triangles to 0).
EXAMPLE
n\k 0 1 2 3 4 5 6 7 ...
0: 1
1: 1 1
2: 1 3 2
3: 1 8 17 10
4: 1 14 65 112 60
5: 1 21 163 567 844 420
6: 1 29 331 1871 5380 7172 3360
7: 1 38 592 4850 22219 55592 67908 30240
...
a(2,2)=a_2(1,2)=A094638(3,3)=1*2=2.
a(2,2) = |s(3,1)| = 2.
a(4,2) = a_2(1,2,5,6) = 1*2+1*5+1*6+2*5+2*6+5*6 = 65.
a(4,2) = 1*(|s(7,5)| - (3*S_3(5,1) + 4*S_4(5,1))) +
3*4*(|s(7,7)| -(3*0 + 4*0)) = 1*(175 -(3*18 + 4*17))
+ 12*1 = 65.
CROSSREFS
KEYWORD
AUTHOR
Wolfdieter Lang, Oct 27 2011
STATUS
approved