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Table[2^(n+1-k) Fibonacci[k], {n, 0, 10}, {k, n+1}]//Flatten (* Harvey P. Dale, Apr 26 2020 *)
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>with(combinat, fibonacci):
>seq(seq(2^(n+1-k)*fibonacci(k), k=1..(n+1)), n=0..10);
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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}. [From _Dennis P. Walsh, _, Dec 21 2011]
Triangle read by rows, A130321 * A127647. A130321 = an infinite lower triangular matrix with powers of 2: (A000079) in every column: (1, 2, 4, 8, ...).
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.
A000045 as the main diagonal and T(n,k)=2^(n+1-k)*F(k) where F(k) is the rest zerosk-th Fibonacci number. [_Dennis Walsh_, Dec 21 2011]
T(n,k)=2^(n+1-k)*F(k) where F(k) is the k-th Fibonacci number. [From Dennis Walsh, Dec 21 2011]
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;
...
...
{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]
[From Dennis Walsh, Dec 21 2011]
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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 consective consecutive integers and there are 2^(n+1-k) subsets of {k+2,...,n+2}. [From Dennis P. Walsh, Dec 21 2011]
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_Gary W. Adamson (qntmpkt(AT)yahoo.com), _, Dec 23 2008