cp's OEIS Frontend

A191386 Number of ascents of length 1 in all dispersed Dyck paths of length n (i.e., in all Motzkin paths of length n with no (1,0) steps at positive heights). An ascent is a maximal sequence of consecutive (1,1)-steps.

%I A191386 #67 Sep 08 2022 08:45:57
%S A191386 0,0,1,2,5,10,23,46,102,204,443,886,1898,3796,8054,16108,33932,67864,
%T A191386 142163,284326,592962,1185924,2464226,4928452,10209620,20419240,
%U A191386 42190558,84381116,173962532,347925064,715908428,1431816856,2941192472,5882384944,12065310083,24130620166
%N A191386 Number of ascents of length 1 in all dispersed Dyck paths of length n (i.e., in all Motzkin paths of length n with no (1,0) steps at positive heights). An ascent is a maximal sequence of consecutive (1,1)-steps.
%C A191386 a(n+2) is the length of a lock-breaking sequence for a lock having buttons 1,2,...,n and a reset button R, and a combination that is any subset of the buttons (the lock opens if the proper combination is pressed after an R). For example, R123R23R31 is a length-10 sequence that unlocks the case of 3 buttons, because each of the 8 subsets occurs somewhere in the sequence between resets. This problem is due to John Guilford. Proof that the shortest sequence has length a(n+2) is due to Dan Velleman and Stan Wagon. - _Stan Wagon_, Feb 17 2019
%H A191386 G. C. Greubel, <a href="/A191386/b191386.txt">Table of n, a(n) for n = 0..1000</a>
%H A191386 F. Blanchet-Sadri, Kun Chen, Kenneth Hawes, <a href="https://arxiv.org/abs/1708.06461">Dyck Words, Lattice Paths, and Abelian Borders</a>, arXiv:1708.06461 [cs.FL], 2017, Conjecture 1.
%H A191386 Kairi Kangro, Mozhgan Pourmoradnasseri, Dirk Oliver Theis, <a href="http://arxiv.org/abs/1603.01422">Short note on the number of 1-ascents in dispersed dyck paths</a>, arXiv:1603.01422 [math.CO], 2016.
%F A191386 G.f.: g(z) = z^2*(1+sqrt(1-4*z^2))/(2*(1-2*z)*sqrt(1-4*z^2)). [The next three formulas follow from this. - _N. J. A. Sloane_, Feb 13 2019]
%F A191386 -(n-2)*a(n) + 2*(n-2)*a(n-1) + 4*(n-3)*a(n-2) - 8*(n-3)*a(n-3) = 0. - _R. J. Mathar_, Jun 14 2016
%F A191386 For n > 1, a(n) = 2^(n - 3) + binomial(n-2, floor(n/2-1))*(n - 1)/2. [See Kangro-Pourmoradnasseri-Theis, first page] - Dan Velleman, Feb 12 2019
%F A191386 a(n) = Sum_{k>=0} k*A191384(n,k).
%F A191386 a(n) ~ 2^(n-5/2)*sqrt(n)/sqrt(Pi) * (1 + sqrt(Pi)/sqrt(2*n)). - _Vaclav Kotesovec_, Mar 21 2014
%e A191386 a(4) = 5 because in HHHH, HHUD, HUDH, UDHH, UDUD, and UUDD we have a total of 0+1+1+1+2+0=5 ascents of length 1.
%p A191386 g := (1/2)*z^2*(1+sqrt(1-4*z^2))/((1-2*z)*sqrt(1-4*z^2)): gser := series(g, z = 0, 38): seq(coeff(gser, z, n), n = 0 .. 35);
%t A191386 CoefficientList[Series[(1/2)*x^2*(1+Sqrt[1-4*x^2])/((1-2*x)*Sqrt[1-4*x^2]), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Mar 21 2014 *)
%o A191386 (PARI) z='z+O('z^40); concat([0,0], Vec(z^2*(1+sqrt(1-4*z^2))/(2*(1-2*z)*sqrt(1-4*z^2)))) \\ _G. C. Greubel_, Mar 26 2017
%o A191386 (Magma) m:=40; R<x>:=PowerSeriesRing(Rationals(), m); [0,0] cat Coefficients(R!( x^2*(1+Sqrt(1-4*x^2))/(2*(1-2*x)*Sqrt(1-4*x^2)) )); // _G. C. Greubel_, Feb 17 2019
%o A191386 (Sage) (x^2*(1+sqrt(1-4*x^2))/(2*(1-2*x)*sqrt(1-4*x^2))).series(x, 40).coefficients(x, sparse=False) # _G. C. Greubel_, Feb 17 2019
%Y A191386 Cf. A191384.
%Y A191386 If the two initial zeros are omitted, we get A323988.
%K A191386 nonn
%O A191386 0,4
%A A191386 _Emeric Deutsch_, Jun 01 2011