|
%I #26 Nov 16 2019 20:08:45
%S 1,1,1,2,2,3,4,1,5,8,3,8,15,8,1,13,28,19,4,21,51,42,13,1,34,92,89,36,
%T 5,55,164,182,91,19,1,89,290,363,216,60,6,144,509,709,489,170,26,1,
%U 233,888,1362,1068,446,92,7,377,1541,2580,2266,1105,288,34,1,610,2662,4830
%N Triangle read by rows: T(n,k) is number of binary words of length n and having k runs of 0's of odd length, 0 <= k <= ceiling(n/2). (A run of 0's is a subsequence of consecutive 0's of maximal length.)
%C Row n has 1+ceiling(n/2) terms.
%C T(n,0) = Fibonacci(n+1) = A000045(n+1).
%C T(n,1) = A029907(n).
%C Sum_{k>=0} k*T(n,k) = A059570(n).
%C Triangle, with zeros included, given by (1,1,-1,0,0,0,0,0,0,0,...) DELTA (1,-1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Dec 07 2011
%C T(n,k) is the number of compositions of n+1 that have exactly k even parts. - _Geoffrey Critzer_, Mar 03 2012
%D I. Goulden and D. Jackson, Combinatorial Enumeration, John Wiley and Sons, 1983, page 54.
%H Alois P. Heinz, <a href="/A119473/b119473.txt">Rows n = 0..200, flattened</a>
%H R. Grimaldi and S. Heubach, <a href="https://www.researchgate.net/publication/220621147_Binary_Strings_Without_Odd_Runs_of_Zeros">Binary strings without odd runs of zeros</a>, Ars Combinatoria 75 (2005), 241-255.
%F G.f.: (1+t*z)/(1-z-z^2-t*z^2).
%F G.f. of column k (k>=1): z^(2*k-1)*(1-z^2)/(1-z-z^2)^(k+1).
%F T(n,k) = T(n-1,k) + T(n-2,k) + T(n-2,k-1). - _Philippe Deléham_, Dec 07 2011
%F Sum_{k=0..n} T(n,k)*x^k = A000045(n+1), A000079(n), A105476(n+1), A159612(n+1), A189732(n+1) for x = 0, 1, 2, 3, 4 respectively. - _Philippe Deléham_, Dec 07 2011
%F G.f.: (1+x*y)*T(0)/2, where T(k) = 1 + 1/(1 - (2*k+1+ x*(1+y))*x/((2*k+2+ x*(1+y))*x + 1/T(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Nov 06 2013
%e T(5,2)=8 because we have 00010, 01000, 01011, 01101, 01110, 10101, 10110 and 11010.
%e T(5,2)=8 because there are 8 compositions of 6 that have 2 even parts: 4+2, 2+4, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2. - _Geoffrey Critzer_, Mar 03 2012
%e Triangle starts:
%e 1;
%e 1, 1;
%e 2, 2;
%e 3, 4, 1;
%e 5, 8, 3;
%e 8, 15, 8, 1;
%e From _Philippe Deléham_, Dec 07 2011: (Start)
%e Triangle (1,1,-1,0,0,0...) DELTA (1,-1,0,0,0,...) begins:
%e 1;
%e 1, 1;
%e 2, 2, 0;
%e 3, 4, 1, 0;
%e 5, 8, 3, 0, 0;
%e 8, 15, 8, 1, 0, 0;
%e 13, 28, 19, 4, 0, 0, 0;
%e 21, 51, 42, 13, 1, 0, 0, 0;
%e 34, 92, 89, 36, 5, 0, 0, 0, 0; ... (End)
%p G:=(1+t*z)/(1-z-z^2-t*z^2): Gser:=simplify(series(G,z=0,18)): P[0]:=1: for n from 1 to 14 do P[n]:=sort(coeff(Gser,z^n)) od: for n from 0 to 14 do seq(coeff(P[n],t,j),j=0..ceil(n/2)) od; # yields sequence in triangular form
%p # second Maple program:
%p b:= proc(n) option remember; local j; if n=0 then 1
%p else []; for j to n do zip((x, y)->x+y, %,
%p [`if`(irem(j, 2)=0, 0, NULL), b(n-j)], 0) od; %[] fi
%p end:
%p T:= n-> b(n+1):
%p seq(T(n), n=0..14); # _Alois P. Heinz_, May 23 2013
%t f[list_] := Select[list, # > 0 &]; nn = 15; a = (x + y x^2)/(1 - x^2); Map[f, Drop[CoefficientList[Series[1/(1 - a), {x, 0, nn}], {x, y}], 1]] // Flatten (* _Geoffrey Critzer_, Mar 03 2012 *)
%Y Cf. A000045, A029907, A059570.
%K nonn,tabf
%O 0,4
%A _Emeric Deutsch_, May 22 2006
|
