%I
%S 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,
%T 128,64,64,48,40,32,26,21,256,128,128,96,80,64,52,42,34,512,256,256,
%U 192,160,128,104,84,68,55
%N Triangle read by rows, A130321 * A127647. Also, number of subsets of [n+2] with consecutive integers that start at k.
%C Row sums = A008466(k2): (1, 3, 8, 19, 43, 94, ...).
%C 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+1k), where F(k) is the kth Fibonacci number, since there are F(k) subsets of {1,...,k2} that contain no consecutive integers and there are 2^(n+1k) subsets of {k+2,...,n+2}. [_Dennis P. Walsh_, Dec 21 2011]
%F Triangle read by rows, A130321 * A127647. A130321 = an infinite lower triangular matrix with powers of 2: (A000079) in every column: (1, 2, 4, 8, ...).
%F A127647 = an infinite lower triangular matrix with the Fibonacci numbers, A000045 as the main diagonal and the rest zeros.
%F T(n,k)=2^(n+1k)*F(k) where F(k) is the kth Fibonacci number. [_Dennis Walsh_, Dec 21 2011]
%e First few rows of the triangle:
%e 1;
%e 2, 1;
%e 4, 2, 2;
%e 8, 4, 4, 3;
%e 16, 8, 8, 6, 5;
%e 32, 16, 16, 12, 10, 8;
%e 64, 32, 32, 24, 20, 16, 13;
%e 128, 64, 64, 48, 40, 32, 26, 21;
%e 256, 128, 128, 96, 80, 64, 52, 42, 34;
%e 512, 256, 256, 192, 160, 128, 104, 84, 68, 55;
%e ...
%e Row 4 = (16, 8, 8, 6, 5) = termwise products of (16, 8, 4, 2, 1) and (1, 1, 2, 3, 5).
%e 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,
%e {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]
%p with(combinat, fibonacci):
%p seq(seq(2^(n+1k)*fibonacci(k),k=1..(n+1)),n=0..10);
%t Table[2^(n+1k) Fibonacci[k],{n,0,10},{k,n+1}]//Flatten (* _Harvey P. Dale_, Apr 26 2020 *)
%Y Cf. A008466, A127647, A130321.
%K nonn,tabl
%O 0,2
%A _Gary W. Adamson_, Dec 23 2008
