|
Search: id:A103633
|
|
|
| A103633 |
|
Triangle read by rows: triangle of repeated stepped binomial coefficients. |
|
+0
3
|
|
| 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 2, 1, 0, 0, 0, 1, 2, 1, 0, 0, 0, 1, 3, 3, 1, 0, 0, 0, 0, 1, 3, 3, 1, 0, 0, 0, 0, 1, 4, 6, 4, 1, 0, 0, 0, 0, 0, 1, 4, 6, 4, 1, 0, 0, 0, 0, 0, 1, 5, 10, 10, 5, 1, 0, 0, 0, 0, 0, 0, 1, 5, 10, 10, 5, 1, 0, 0, 0, 0, 0, 0, 1, 6, 15, 20, 15, 6, 1, 0, 0, 0, 0, 0, 0, 0, 1, 6, 15
(list; table; graph; listen)
|
|
|
OFFSET
|
0,14
|
|
|
COMMENT
|
Row sums are sum{k=0..n, binomial(floor(n/2),n-k)}=(1,1,2,2,4,4,...). Diagonal sums have g.f. (1+x^2)/(1-x^3-x^4) (see A079398). Matrix inverse of the signed triangle (-1)^(n-k)T(n,k) is A103631. Matrix inverse of T(n,k) is the alternating signed version of A103631.
Triangle T(n,k), 0<=k<=n, read by rows, given by [0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ....] DELTA [1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938 . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 08 2005
|
|
FORMULA
|
Number triangle T(n, k)=binomial(floor(n/2), n-k)
Sum_{n, n>=0} T(n, k) = A000045(k+2) = Fib(k+2) . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 08 2005
Sum_{k, 0<=k<=n}T(n,k)=2^[n/2]=A016116(n) . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Dec 03 2006
|
|
EXAMPLE
|
Rows begin {1}, {0,1}, {0,1,1}, {0,0,1,1}, {0,0,0,1,2,1},...
|
|
CROSSREFS
|
Sequence in context: A064662 A024944 A117907 this_sequence A026821 A039964 A035172
Adjacent sequences: A103630 A103631 A103632 this_sequence A103634 A103635 A103636
|
|
KEYWORD
|
easy,nonn,tabl
|
|
AUTHOR
|
Paul Barry (pbarry(AT)wit.ie), Feb 11 2005
|
|
|
Search completed in 0.003 seconds
|
