|
Search: id:A105260
|
|
|
| A105260 |
|
Triangle read by rows: T(n,k)=C(2n-2k,k), n>=0, 0<=k<=floor(2n/3). |
|
+0 1
|
|
| 1, 1, 1, 2, 1, 4, 1, 1, 6, 6, 1, 8, 15, 4, 1, 10, 28, 20, 1, 1, 12, 45, 56, 15, 1, 14, 66, 120, 70, 6, 1, 16, 91, 220, 210, 56, 1, 1, 18, 120, 364, 495, 252, 28, 1, 20, 153, 560, 1001, 792, 210, 8, 1, 22, 190, 816, 1820, 2002, 924, 120, 1, 1, 24, 231, 1140, 3060, 4368, 3003
(list; graph; listen)
|
|
|
OFFSET
|
0,4
|
|
|
REFERENCES
|
E. Deutsch, Math. Magazine, vol. 75, No. 3, 2002, p. 228, problem 1623.
|
|
FORMULA
|
T(n, k)=C(2n-2k, k), n>=0, 0<=k<=floor(2n/3). G.f.=1/[1-z(1+tz)^2].
|
|
EXAMPLE
|
Triangle begins:
1;
1;
1,2;
1,4,1;
1,6,6;
1,8,15,4;
Row n contains 1+floor(2n/3) terms.
|
|
MAPLE
|
T:=(n, k)->binomial(2*n-2*k, k): for n from 0 to 14 do seq(T(n, k), k=0..floor(2*n/3)) od; # yields sequence in triangular form
|
|
CROSSREFS
|
Row sums yield A002478.
Sequence in context: A124844 A133934 A055327 this_sequence A099510 A137633 A066633
Adjacent sequences: A105257 A105258 A105259 this_sequence A105261 A105262 A105263
|
|
KEYWORD
|
nonn,tabf
|
|
AUTHOR
|
Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 14 2005
|
|
|
Search completed in 0.002 seconds
|