|
Search: id:A059481
|
|
|
| A059481 |
|
Triangle read by rows: T(n,k) = number of ways to distribute k identical objects in n distinct containers; containers may be left empty. |
|
+0 7
|
|
| 1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 462, 1, 7, 28, 84, 210, 462, 924, 1716, 1, 8, 36, 120, 330, 792, 1716, 3432, 6435, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 24310, 1, 10, 55, 220, 715, 2002, 5005, 11440
(list; table; graph; listen)
|
|
|
OFFSET
|
1,5
|
|
|
COMMENT
|
Coefficients of Faber polynomials for function x^2/(x-1). - Michael Somos, Sep 09 2003
|
|
REFERENCES
|
R. Grimaldi, Discrete and Combinatorial Mathematics, Addison-Wesley, 4:th edition, chapter 1.4.
|
|
FORMULA
|
T(n, k)=binomial(n+k-1, k), but triangle includes only n>=k>=0.
|
|
EXAMPLE
|
1; 1,1; 1,2,3; 1,3,6,10; 1,4,10,20,35; ...
T(3,3) = 10 because the ways to distribute the 3 objects into the three containers are: (3,0,0) (0,3,0) (0,0,3) (2,1,0) (1,2,0) (2,0,1) (1,0,2) (0,1,2) (0,2,1) (1,1,1), for a total of 10 possibilities.
T(3,3)=10 since (x^2/(x-1))^3 = (x+1+1/x+O(1/x^2))^3 = x^3+3x^2+6x+10+O(x).
|
|
MAPLE
|
for n from 0 to 20 do for m from 0 to n do printf(`%d, `, binomial(n+m-1, m)) od:od:
|
|
PROGRAM
|
(PARI) T(n, k)=if(k<0|k>n, 0, binomial(n+k-1, k))
(PARI) T(n, k)=if(n<0, 0, polcoeff(Pol(((1/(x-x^2)+x*O(x^n))^n+O(x))*x^n), k))
|
|
CROSSREFS
|
Take Pascal's triangle A007318, delete entries to the right of a vertical line just right of center, then scan the diagonals.
For a signed version of this triangle see A027555.
Row sums give A000984.
Sequence in context: A111808 A081422 A027555 this_sequence A158498 A113592 A136555
Adjacent sequences: A059478 A059479 A059480 this_sequence A059482 A059483 A059484
|
|
KEYWORD
|
easy,nice,nonn,tabl
|
|
AUTHOR
|
Fabian Rothelius (fabian.rothelius(AT)telia.com), Feb 04 2001
|
|
EXTENSIONS
|
More terms from James A. Sellers (sellersj(AT)math.psu.edu), Feb 07 2001
|
|
|
Search completed in 0.002 seconds
|