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Number of Dumont permutations of the first kind of length 2n avoiding the patterns 1423 and 4132.
3

%I #39 Jan 20 2024 09:17:59

%S 1,1,3,12,52,232,1049,4777,21845,100159,460023,2115350,9735205,

%T 44829766,206526972,951759621,4387156587,20226421380,93264500832,

%U 430091815527,1983549213861,9148582037193,42197572190160,194643215702835

%N Number of Dumont permutations of the first kind of length 2n avoiding the patterns 1423 and 4132.

%C [1, 3, 12, 52, 232, ...] is INVERT transform of [1, 2, 27, 108, 440, ...] A026726. - _Michael Somos_, Apr 15 2012

%C HANKEL transform of sequence and the sequence omitting a(0) is the odd and even bisections of Fibonacci numbers respectively. This is the unique sequence with that property. - _Michael Somos_, Apr 15 2012

%C Bisection (even part) of A224747. - _Alois P. Heinz_, Jul 29 2013

%H Alois P. Heinz, <a href="/A125187/b125187.txt">Table of n, a(n) for n = 0..1000</a>

%H Cyril Banderier, Michael Wallner, <a href="https://arxiv.org/abs/1707.01931">Lattice paths with catastrophes</a>, arXiv:1707.01931 [math.CO], 2017.

%H Paul Barry, <a href="https://arxiv.org/abs/1912.01124">A Note on Riordan Arrays with Catalan Halves</a>, arXiv:1912.01124 [math.CO], 2019.

%H A. Burstein, <a href="http://arxiv.org/abs/math/0402378">Restricted Dumont permutations</a>, arXiv:math/0402378 [math.CO], 2004

%H A. Burstein, <a href="http://dx.doi.org/10.1007/s00026-005-0256-4">Restricted Dumont permutations</a>, Annals of Combinatorics, 9, 2005, 269-280 (Theorem 3.12).

%F G.f.: [2-(1+x)C(x)]/[2-x-(1+x)C(x)], where C(x)=(1-sqrt(1-4x))/(2x) is the Catalan function.

%F From _Gary W. Adamson_, Jul 11 2011: (Start)

%F a(n) = upper left term in M^n, where M is an infinite square production matrix in which two columns of (1,2,3,...) are prepended to an infinite lower triangular matrix of all 1's and the rest zeros, as follows:

%F 1, 1, 0, 0, 0, 0, ...

%F 2, 2, 1, 0, 0, 0, ...

%F 3, 3, 1, 1, 0, 0, ...

%F 4, 4, 1, 1, 1, 0, ...

%F 5, 5, 1, 1, 1, 1, ...

%F ... (End)

%F Given g.f. A(x), then 0 = A(x)^2 * (x^3 - 2*x^2 + 5*x - 1) + A(x) *(x^2 - 9*x + 2) + (x^2 + 4*x -1). - _Michael Somos_, Jan 14 2014

%F 0 = a(n)*(16*a(n+1) +6*a(n+2) -14*a(n+3) +210*a(n+4) -128*a(n+5) +18*a(n+6)) +a(n+1)*(-46*a(n+1) +143*a(n+2) -173*a(n+3) -283*a(n+4) +202*a(n+5) -29*a(n+6)) +a(n+2)*(-63*a(n+2) +386*a(n+3) +765*a(n+4) -529*a(n+5) +75*a(n+6)) +a(n+3)*(-559*a(n+3) +509*a(n+4) -149*a(n+5) +19*a(n+6)) +a(n+4)*(-108*a(n+4) +71*a(n+5) -12*a(n+6)) +a(n+5)*(-4*a(n+5) +a(n+6)). - _Michael Somos_, Jan 14 2014

%F G.f.: ( 2 - 9*x + x^2 + (x + x^2) * sqrt(1 - 4*x) ) / (2 - 10*x + 4*x^2 - 2*x^3). - _Michael Somos_, Apr 15 2012

%F G.f. = (1 - 3*y + y^2) / (1 - 4*y + 3*y^2 - y^3) = 1 / (1 - y / (1 - y / (1 - 2*y / (1 + y / (2 - y))))) where y = (1 - sqrt(1 - 4*x)) / 2. - _Michael Somos_, Apr 12 2012

%F D-finite with recurrence (-n+1)*a(n) +4*(2*n-3)*a(n-1) +(-13*n+19)*a(n-2) +(-13*n+75)*a(n-3) +(5*n-29)*a(n-4) +2*(-2*n+9)*a(n-5)=0. - _R. J. Mathar_, Jul 27 2013

%e G.f. = 1 + x + 3*x^2 + 12*x^3 + 52*x^4 + 232*x^5 + 1049*x^6 + 4777*x^7 + 21845*x^8 + ...

%p C:=(1-sqrt(1-4*x))/2/x: G:=(2-(1+x)*C)/(2-x-(1+x)*C): Gser:=series(G,x=0,30): seq(coeff(Gser,x,n),n=0..26);

%t a[ n_] := SeriesCoefficient[ (2 - 9 x + x^2 + (x + x^2) Sqrt[1 - 4 x]) / (2 (1 - 5 x + 2 x^2 - x^3)), {x, 0, n}]; (* _Michael Somos_, Jan 14 2014 *)

%o (PARI) {a(n) = if( n<0, 0, polcoeff( (2 - 9*x + x^2 + (x + x^2) * sqrt(1 - 4*x + x * O(x^n)) ) / (2 * (1 - 5*x + 2*x^2 - x^3)), n))}; /* _Michael Somos_, Jan 14 2014 */

%Y Cf. A125188, A224747.

%K nonn

%O 0,3

%A _Emeric Deutsch_, Dec 19 2006