

A109435


Triangle read by rows: T(n,m) = number of binary numbers n digits long, which have m 0's as a substring.


2



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
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OFFSET

0,2


COMMENTS

Column 0 is A000079, column 2 is A000225, column 3 is A008466, column 4 is A050231
Column 5 is A050232, column 6 is A050233, the last column is A001792.
A050227 with a leading column of powers of 2.  R. J. Mathar, Mar 25 2014


LINKS

Table of n, a(n) for n=0..70.


FORMULA

G.f. for column m: x^m/( (1  Sum_{k=1..m} x^k)*(12*x) ).  Geoffrey Critzer, Jan 07 2014


EXAMPLE

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


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/(1Sum[x^k, {k, 1, m}])/(12x), {x, 0, nn}], x], {m, 0, nn}]]]//Grid (* Geoffrey Critzer, Jan 07 2014 *)


CROSSREFS

Cf. A109433, A001792, A109436.
Sequence in context: A134626 A347633 A115450 * A134392 A048483 A276562
Adjacent sequences: A109432 A109433 A109434 * A109436 A109437 A109438


KEYWORD

base,nonn,tabl


AUTHOR

Robert G. Wilson v, Jun 28 2005


STATUS

approved



