%I #35 Sep 29 2024 03:50:05
%S 0,0,0,0,1,2,5,12,28,62,136,294,628,1328,2787,5810,12043,24840,51016,
%T 104380,212848,432732,877400,1774672,3581605,7213746,14502449,
%U 29106100,58323844,116702074,233199000,465405058,927744428,1847359520,3674769991
%N Number of binary sequences of length n containing exactly one subsequence 0000.
%C Column 1 of A118897.
%H Michael De Vlieger, <a href="/A118898/b118898.txt">Table of n, a(n) for n = 0..3501</a>
%H Omar Khadir, László Németh, and László Szalay, <a href="https://doi.org/10.1007/s00025-024-02284-3">Tiling of dominoes with ranked colors</a>, Results in Math. (2024) Vol. 79, Art. No. 253. See p. 2.
%H László Németh and László Szalay, <a href="https://arxiv.org/abs/2408.12196">Explicit solution of system of two higher-order recurrences</a>, arXiv:2408.12196 [math.NT], 2024. See p. 10.
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,0,-1,-4,-3,-2,-1).
%F G.f.: z^4/(1-z-z^2-z^3-z^4)^2.
%F From _Bobby Milazzo_, Aug 30 2009: (Start)
%F a(1)=0,a(2)=0,a(3)=0,a(4)=1,a(5)=2,a(6)=5,a(7)=12,a(8)=28
%F a(n) = 2a(n-1)+a(n-2)-a(n-4)-4a(n-5)-3a(n-6)-2a(n-7)-a(n-8). (End)
%e a(6)=5 because we have 000010,000011,010000,100001 and 110000.
%e G.f. = x^4 + 2*x^5 + 5*x^6 + 12*x^7 + 28*x^8 + 62*x^9 + ... - _Zerinvary Lajos_, Jun 02 2009
%p g:=z^4/(1-z-z^2-z^3-z^4)^2: gser:=series(g,z=0,40): seq(coeff(gser,z,n),n=0..37);
%t RecurrenceTable[{a[1]==0,a[2]==0,a[3]==0,a[4]==1,a[5]==2,a[6]==5, a[7]==12,a[8]==28,a[n]==2a[n-1]+a[n-2]-a[n-4]-4a[n-5]-3a[n-6]-2a[n-7]-a[n-8]},a,{n,9,50}] (* _Bobby Milazzo_, Aug 30 2009 *)
%t LinearRecurrence[{2,1,0,-1,-4,-3,-2,-1},{0,0,0,0,1,2,5,12},50] (* _Harvey P. Dale_, Aug 01 2012 *)
%o (Sage) taylor( mul(x/(1-x-x^2-x^3-x^4)^2 for i in range(1,2)),x,0,31)# _Zerinvary Lajos_, Jun 02 2009
%Y Cf. A118897.
%K nonn,easy
%O 0,6
%A _Emeric Deutsch_, May 04 2006