A153281 - OEIS

%I #17 Apr 26 2020 11:39:55

%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(k-2): (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+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]

%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+1-k)*F(k) where F(k) is the k-th 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+1-k)*fibonacci(k),k=1..(n+1)),n=0..10);

%t Table[2^(n+1-k) 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