A109433 Triangle read by rows: T(n,m) = number of binary numbers n+1 digits long which have m 1's as a substring.

1, 2, 1, 4, 2, 1, 8, 5, 2, 1, 16, 11, 5, 2, 1, 32, 24, 12, 5, 2, 1, 64, 51, 27, 12, 5, 2, 1, 128, 107, 60, 28, 12, 5, 2, 1, 256, 222, 131, 63, 28, 12, 5, 2, 1, 512, 457, 282, 140, 64, 28, 12, 5, 2, 1, 1024, 935, 601, 307, 143, 64, 28, 12, 5, 2, 1, 2048, 1904, 1270, 666, 316, 144

Offset: 0

Author

Robert G. Wilson v, Jun 27 2005

Keywords

• base
• nonn
• tabl

Examples

			T(4,2)=11 because of the sixteen binary digits which are 5 long, {10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111}, 11 have "11" as a substring.
Triangle begins:
n\m
0 1 0 0 0 0 0 0 0 0 0
1 2 1 0 0 0 0 0 0 0 0
2 4 2 1 0 0 0 0 0 0 0
3 8 5 2 1 0 0 0 0 0 0
4 16 11 5 2 1 0 0 0 0 0
5 32 24 12 5 2 1 0 0 0 0
		

Crossrefs

First column = A000079 = Powers of 2, the second column = A027934 = number of compositions of n with at least one even part and the last column = A045623 = number of 1's in all compositions of n+1.

Cf. A000079, A027934, A045623, A109434, A109435.

Programs

Mathematica

T[n_, m_] := Length[ Select[ StringPosition[ #, ToString[(10^m - 1)/9]] & /@ Table[ ToString[ FromDigits[ IntegerDigits[i, 2]]], {i, 2^n, 2^(n + 1) - 1}], # != {} &]]; Flatten[ Table[ T[n, m], {n, 0, 11}, {m, n + 1}]]