Continuing we get $ $ G_ { HT } ( omega ) = G_ { HH } ( omega ) \frac { z } { 1-z }. $ $
furthermore $ $ G_ { TT } ( omega ) = \frac { omega } { 1-z } \sum_ { k=0 } ^\infty \left ( z\frac { 1-z^ { q } } { 1-z } \frac { z } { 1-z } \right ) ^k. $ $
ultimately we have $ $ G_ { TH } ( z ) = G_ { TT } ( z ) z\frac { 1-z^ { q } } { 1-z }. $ $
The sum term is $ $ \frac { 1 } { 1-z^2 ( 1-z^q ) / ( 1-z ) ^2 } = \frac { 1-2z+z^2 } { 1-2z+z^2-z^2 ( 1-z^q ) } = \frac { 1-2z+z^2 } { 1-2z+z^ { q+2 } }. $ $
The gene on this is $ $ z\frac { 1-z^ { q } } { 1-z } \left ( 1+\frac { z } { 1-z } \right ) + \frac { z } { 1-z } \left ( 1+z\frac { 1-z^ { q } } { 1-z } \right ) $ $ which is $ $ z\frac { 1-z^ { q } } { ( 1-z ) ^2 } + \frac { z } { ( 1-z ) ^2 } ( 1-z^ { q+1 } ) = \frac { 2z-z^ { q+1 } -z^ { q+2 } } { ( 1-z ) ^2 }. $ $
Multiplying we obtain the generating function $ $ G_q ( z ) = \frac { 2z-z^ { q+1 } -z^ { q+2 } } { 1-2z+z^ { q+2 } }. $ $
It follows that the expectation time $ 2^n $ is given by
Read more: Mini Coin Purse – Free Crochet Pattern
$ $ [ z^n ] \left ( 0\times G_0 ( z ) + \sum_ { q=1 } ^n q ( G_q ( z ) -G_ { q-1 } ( omega ) ) \right ). $ $
The sum simplifies to $ $ \sum_ { q=1 } ^n q G_q ( z ) – \sum_ { q=0 } ^ { n-1 } ( q+1 ) G_q ( z ) = \sum_ { q=0 } ^n q G_q ( z ) – \sum_ { q=0 } ^ { n-1 } ( q+1 ) G_q ( omega ) \\ = normality G_n ( omega ) – \sum_ { q=0 } ^ { n-1 } G_q ( z ). $ $
and hence the expectation is $ $ \frac { 1 } { 2^n } [ z^n ] \left ( nitrogen G_n ( omega ) – \sum_ { q=0 } ^ { n-1 } G_q ( omega ) \right ). $ $
This gives the sequence $ $ 1/2,1, { \frac { 11 } { 8 } }, { \frac { 27 } { 16 } }, { \frac { 31 } { 16 } }, { \frac { 69 } { 32 } }, { \frac { 75 } { 32 } }, { \frac { 643 } { 256 } }, { \frac { 1363 } { 512 } }, { \frac { 1433 } { 512 } }, \ldots $ $
Multiplying by $ 2^n $ we obtain $ $ 1, 4, 11, 27, 62, 138, 300, 643, 1363, 2866, \ldots $ $ which is OEIS A119706 where the above calculation is confirmed .
The following Maple code can be used to explore these generating functions. The procedure v computes the generating function of the maximal run distance of a string of $ n $ bits by total count. The routine w computes it from the generating serve $ G_q ( omega ). $
v := proc(n) option remember; local gf, k, d, mxrun, len; gf := 0; for k from 2^n to 2^(n+1)-1 do d := convert(k, base, 2); mxrun := 0; for pos to n do if d[pos] = 1 then len := 1; pos := pos+1; while pos <= n do if d[pos] = 1 then len := len+1; pos := pos+1; else break; fi; od; if len>mxrun then mxrun := len; fi; fi; od; gf := gf + z^mxrun; od; gf; end; G := q -> (2*z-z^(q+1)-z^(q+2))/(1-2*z+z^(q+2)); w := proc(n) option remember; local gf, mxrun; gf := 1; for mxrun to n do gf := gf + coeftayl(G(mxrun)-G(mxrun-1), z=0, n)*z^mxrun; od; gf; end; X := n -> coeftayl(n*G(n)-add(G(q), q=0..n-1), z=0, n)/2^n;
here are two examples .
> v(4); 4 3 2 z + 2 z + 5 z + 7 z + 1 > w(4); 4 3 2 z + 2 z + 5 z + 7 z + 1
Addendum. Responding to the question of the OP, the maximal test length distribution for $ n=50 $ is
> w(50); 50 49 48 47 46 45 44 43 42 z + 2 z + 5 z + 12 z + 28 z + 64 z + 144 z + 320 z + 704 z 41 40 39 38 37 36 + 1536 z + 3328 z + 7168 z + 15360 z + 32768 z + 69632 z 35 34 33 32 31 + 147456 z + 311296 z + 655360 z + 1376256 z + 2883584 z 30 29 28 27 26 + 6029312 z + 12582912 z + 26214400 z + 54525952 z + 113246208 z 25 24 23 22 + 234881024 z + 486539259 z + 1006632909 z + 2080374408 z 21 20 19 18 + 4294964912 z + 8858356224 z + 18253535488 z + 37580568576 z 17 16 15 14 + 77307408384 z + 158903894017 z + 326369607799 z + 669786836360 z 13 12 11 + 1373319005440 z + 2812533538048 z + 5749650288420 z 10 9 8 + 11716183298140 z + 23723022576779 z + 47402584528885 z 7 6 5 + 92066138963408 z + 168050756947888 z + 267156803852044 z 4 3 2 + 310228979841119 z + 174887581402185 z + 19394019617001 z + 32951280098 z + 1
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