A109435 Triangle read by rows: T(n,m) = number of binary numbers n digits long, which have m 0's as a substring.
1, 2, 1, 4, 3, 1, 8, 7, 3, 1, 16, 15, 8, 3, 1, 32, 31, 19, 8, 3, 1, 64, 63, 43, 20, 8, 3, 1, 128, 127, 94, 47, 20, 8, 3, 1, 256, 255, 201, 107, 48, 20, 8, 3, 1, 512, 511, 423, 238, 111, 48, 20, 8, 3, 1, 1024, 1023, 880, 520, 251, 112, 48, 20, 8, 3, 1, 2048, 2047, 1815, 1121, 558
Offset: 0
Examples
Triangle begins: n\m_0__1__2__3__4__5 0| 1 0 0 0 0 0 1| 2 1 0 0 0 0 2| 4 3 1 0 0 0 3| 8 7 3 1 0 0 4| 16 15 8 3 1 0 5| 32 31 19 8 3 1 T(5,3)=8 because there are 8 length 5 binary words that contain 000 as a contiguous substring: 00000, 00001, 00010, 00011, 01000, 10000, 10001, 11000. - _Geoffrey Critzer_, Jan 07 2014
Programs
-
Maple
A109435 := proc(n,k) option remember ; if n< k then 0; elif n = k then 1; else 2*procname(n-1,k)+2^(n-1-k)-procname(n-1-k,k) ; end if; end proc: seq(seq( A109435(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Jun 05 2025
-
Mathematica
T[n_, m_] := Length[ Select[ StringPosition[ #, StringDrop[ ToString[10^m], 1]] & /@ Table[ ToString[ FromDigits[ IntegerDigits[i, 2]]], {i, 2^n, 2^(n + 1) - 1}], # != {} &]]; Flatten[ Table[ T[n, m], {n, 0, 11}, {m, 0, n}]] nn=15;Map[Select[#,#>0&]&,Transpose[Table[CoefficientList[Series[x^m/(1-Sum[x^k,{k,1,m}])/(1-2x),{x,0,nn}],x],{m,0,nn}]]]//Grid (* Geoffrey Critzer, Jan 07 2014 *)
Formula
G.f. for column m: x^m/( (1 - Sum_{k=1..m} x^k)*(1-2*x) ). - Geoffrey Critzer, Jan 07 2014
Comments