A228612 Number of (possibly overlapping) occurrences of the subword given by the binary expansion of n in all binary words of length n.

0, 1, 1, 4, 4, 12, 32, 80, 80, 192, 448, 1024, 2304, 5120, 11264, 24576, 24576, 53248, 114688, 245760, 524288, 1114112, 2359296, 4980736, 10485760, 22020096, 46137344, 96468992, 201326592, 419430400, 872415232, 1811939328, 1811939328, 3758096384, 7784628224

Offset: 0

Author

Alois P. Heinz, Dec 18 2013

Keywords

• nonn

Comments

a(2^n) = a(2^n-1) for n>0.

Examples

			a(3) = 4 because we have one subword 11 in each of 011, 110 and two overlapping occurrences of 11 in 111.
a(4) = 4 because we have one subword 100 in each of 0100, 1000, 1001, 1100 and no other occurrences in binary words of length 4.
a(5) = 12 because we have one subword 101 in each of 00101, 01010, 01011, 01101, 10100, 10110, 10111, 11010, 11011, 11101 and two overlapping occurrences of 101 in 10101.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A233940.

Formula

a(n) = Sum_{k>0} k*A233940(n,k).