|
|
A121547
|
|
Fourth slice along the 1-2-plane in the cube a(m,n,o) = a(m-1,n,o) + a(m,n-1,o) + a(m,n,o-1) for which the first slice is Pascal's triangle (slice read by antidiagonals).
|
|
2
|
|
|
0, 0, 1, 0, 4, 4, 0, 10, 20, 10, 0, 20, 60, 60, 20, 0, 35, 140, 210, 140, 35, 0, 56, 280, 560, 560, 280, 56, 0, 84, 504, 1260, 1680, 1260, 504, 84, 0, 120, 840, 2520, 4200, 4200, 2520, 840, 120, 0, 165, 1320, 4620, 9240, 11550, 9240, 4620, 1320, 165, 0, 220, 1980, 7920, 18480, 27720, 27720, 18480, 7920, 1980, 220
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,5
|
|
COMMENTS
|
Essentially the same as triangle A178820 with an additional column of zeros at the left border. - Georg Fischer, Jul 31 2023
|
|
LINKS
|
|
|
FORMULA
|
a(m-1,n,o) + a(m,n-1,o) + a(m,n,o-1) with initialization values a(1,0,0) = 1 and a(m<>1=0, n>=0, 0>=o) = 0.
|
|
EXAMPLE
|
The second row is 1, 4, 10, 20, 35, 56, 84, 120, 165, 220 = A000292, i.e., Tetrahedral (or pyramidal) numbers: binomial(n+2,3) = n(n+1)(n+2)/6 (core).
The third row is 4, 20, 60, 140, 280, 504, 840, 1320, 1980, 2860 = A033488 = n*(n+1)*(n+2)*(n+3)/6.
The main diagonal is 0, 4, 60, 560, 4200, 27720, 168168, 960960, 5250960, 27713400 = {0} U A002803*4.
Triangle starts:
0
0, 1
0, 4, 4
0, 10, 20, 10
0, 20, 60, 60, 20
0, 35, 140, 210, 140, 35
0, 56, 280, 560, 560, 280, 56
0, 84, 504, 1260, 1680, 1260, 504, 84
|
|
MAPLE
|
T:=(n, k)->binomial(n+3, 3)*binomial(n, k): seq(print(seq(T(n-1, k-1), k=0..n)), n=0..10); # Georg Fischer, Jul 31 2023
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
EXTENSIONS
|
a(55)-a(56) corrected and more terms from Georg Fischer, Jul 31 2023
|
|
STATUS
|
approved
|
|
|
|