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%I #17 Mar 25 2014 16:05:10
%S 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,
%T 127,94,47,20,8,3,1,256,255,201,107,48,20,8,3,1,512,511,423,238,111,
%U 48,20,8,3,1,1024,1023,880,520,251,112,48,20,8,3,1,2048,2047,1815,1121,558
%N Triangle read by rows: T(n,m) = number of binary numbers n digits long, which have m 0's as a substring.
%C Column 0 is A000079, column 2 is A000225, column 3 is A008466, column 4 is A050231
%C Column 5 is A050232, column 6 is A050233, the last column is A001792.
%C A050227 with a leading column of powers of 2. - _R. J. Mathar_, Mar 25 2014
%F G.f. for column m: x^m/( (1 - Sum_{k=1..m} x^k)*(1-2*x) ). - _Geoffrey Critzer_, Jan 07 2014
%e Triangle begins:
%e n\m_0__1__2__3__4__5
%e 0| 1 0 0 0 0 0
%e 1| 2 1 0 0 0 0
%e 2| 4 3 1 0 0 0
%e 3| 8 7 3 1 0 0
%e 4| 16 15 8 3 1 0
%e 5| 32 31 19 8 3 1
%e 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
%t 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}]]
%t 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 *)
%Y Cf. A109433, A001792, A109436.
%K base,nonn,tabl
%O 0,2
%A _Robert G. Wilson v_, Jun 28 2005
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