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A103633
Triangle read by rows: triangle of repeated stepped binomial coefficients.
2
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
OFFSET
0,14
COMMENTS
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 Deléham, Oct 08 2005
FORMULA
Number triangle T(n, k) = binomial(floor(n/2), n-k).
Sum_{n>=0} T(n, k) = A000045(k+2) = Fibonacci(k+2). - Philippe Deléham, Oct 08 2005
Sum_{k=0..n} T(n,k) = 2^floor(n/2) = A016116(n). - Philippe Deléham, Dec 03 2006
G.f.: (1+x*y)/(1-x^2*y-x^2*y^2). - Philippe Deléham, Nov 10 2013
T(n,k) = T(n-2,k-1) + T(n-2,k-2) for n > 2, T(0,0) = T(,1) = T(2,1) = T(2,2) = 1, T(1,0) = T(2,0) = 0, T(n,k) = 0 if k > n or if k < 0. - Philippe Deléham, Nov 10 2013
EXAMPLE
Triangle begins:
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; ...
CROSSREFS
Sequence in context: A281244 A284585 A280456 * A350847 A026821 A377334
KEYWORD
easy,nonn,tabl
AUTHOR
Paul Barry, Feb 11 2005
STATUS
approved