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A153281
Triangle read by rows, A130321 * A127647. Also, number of subsets of [n+2] with consecutive integers that start at k.
1
1, 2, 1, 4, 2, 2, 8, 4, 4, 3, 16, 8, 8, 6, 5, 32, 16, 16, 12, 10, 8, 64, 32, 32, 24, 20, 16, 13, 128, 64, 64, 48, 40, 32, 26, 21, 256, 128, 128, 96, 80, 64, 52, 42, 34, 512, 256, 256, 192, 160, 128, 104, 84, 68, 55
OFFSET
0,2
COMMENTS
Row sums = A008466(k-2): (1, 3, 8, 19, 43, 94, ...).
T(n,k) is the number of subsets of {1,...,n+2} that contain consecutive integers and that have k as the first integer in the first consecutive string. (See the example below.) Hence rows sums of T(n,k) give the number of subsets of {1,...,n+2} that contain consecutive integers. Also, T(n,k) = F(k)*2^(n+1-k), where F(k) is the k-th Fibonacci number, since there are F(k) subsets of {1,...,k-2} that contain no consecutive integers and there are 2^(n+1-k) subsets of {k+2,...,n+2}. [Dennis P. Walsh, Dec 21 2011]
FORMULA
Triangle read by rows, A130321 * A127647. A130321 = an infinite lower triangular matrix with powers of 2: (A000079) in every column: (1, 2, 4, 8, ...).
A127647 = an infinite lower triangular matrix with the Fibonacci numbers, A000045 as the main diagonal and the rest zeros.
T(n,k)=2^(n+1-k)*F(k) where F(k) is the k-th Fibonacci number. [_Dennis Walsh_, Dec 21 2011]
EXAMPLE
First few rows of the triangle:
1;
2, 1;
4, 2, 2;
8, 4, 4, 3;
16, 8, 8, 6, 5;
32, 16, 16, 12, 10, 8;
64, 32, 32, 24, 20, 16, 13;
128, 64, 64, 48, 40, 32, 26, 21;
256, 128, 128, 96, 80, 64, 52, 42, 34;
512, 256, 256, 192, 160, 128, 104, 84, 68, 55;
...
Row 4 = (16, 8, 8, 6, 5) = termwise products of (16, 8, 4, 2, 1) and (1, 1, 2, 3, 5).
For n=5 and k=3, T(5,3)=16 since there are 16 subsets of {1,2,3,4,5,6,7} containing consecutive integers with 3 as the first integer in the first consecutive string, namely,
{1,3,4}, {1,3,4,5}, {1,3,4,6}, {1,3,4,7}, {1,3,4,5,6}, {1,3,4,5,7}, {1,3,4,6,7}, {1,3,4,5,6,7}, {3,4}, {3,4,5}, {3,4,6}, {3,4,7}, {3,4,5,6}, {3,4,5,7}, {3,4,6,7}, and {3,4,5,6,7}. [Dennis P. Walsh, Dec 21 2011]
MAPLE
with(combinat, fibonacci):
seq(seq(2^(n+1-k)*fibonacci(k), k=1..(n+1)), n=0..10);
MATHEMATICA
Table[2^(n+1-k) Fibonacci[k], {n, 0, 10}, {k, n+1}]//Flatten (* Harvey P. Dale, Apr 26 2020 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Gary W. Adamson, Dec 23 2008
STATUS
approved