A131454 2 up, 2 down, ..., 2 up, 2 down, 2 up permutations of length 4n+3.

1, 71, 45541, 120686411, 908138776681, 15611712012050351, 531909061958526321421, 32491881630252866646683891, 3302814916156503291298772711761, 527393971346575736206847604137659031, 126355819963625435928020023737689391659701

Offset: 0

Author

Peter Bala, Jul 13 2007

Keywords

• easy
• nonn

Comments

Bisection of A005981. The entries listed above suggest various congruences for a(n): a(n) = 1 (mod 10), a(n) = 1 + 70*n (mod 100), a(n) = 1 + 70*n + 200*n(n-1) (mod 1000). Are these congruences true for all n? For an arbitrary integer m, the sequence a(n) taken modulo m may eventually become periodic. Compare with A081727.

Examples

			(1 4 5 3 2 6 7) is a 2 up, 2 down, 2 up permutation - one of the seventy-one permutations of this type in the symmetric group on 7 letters.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..50
• Christopher R. H. Hanusa, Alejandro H. Morales, and Martha Yip, Column convex matrices, G-cyclic orders, and flow polytopes, arXiv:2107.07326 [math.CO], 2021.
• L. Olivier, Bemerkungen über eine Art von Functionen, welche ähnliche Eigenschaften haben, wie der Cosinus und Sinus, J. Reine Angew. Math. 2 (1827), 243-251.
• H. Prodinger and T. A. Tshifhumulo, On q-Olivier Functions, Annals of Combinatorics 6 (2002), 181-194.
• B. Shapiro and A. Vainshtein, Counting real rational functions with all real critical values, arXiv:math/0209062 [math.AG], 2002.
• B. Shapiro and A. Vainshtein, Counting real rational functions with all real critical values, Moscow Math. J. 3 (2003), 647-659.

Crossrefs

Cf. A000111, A005981, A058257, A131453, A131455.

Programs

Maple

g:=(sinh(x)-sin(x))/(cos(x)*cosh(x)+1): series(%,x,44):
seq(n!*coeff(%,x,n),n=3..45,4); # Peter Luschny, Feb 07 2017

Mathematica

Table[(CoefficientList[Series[(-Sin[x] + Sinh[x]) / (1 + Cos[x]*Cosh[x]), {x, 0, 60}], x] * Range[0, 59]!)[[n]], {n, 4, 60, 4}] (* Vaclav Kotesovec, Sep 09 2014 *)

Formula

E.g.f.: Sum_{n>=0} a(n)*(x^(4n+3))/(4n+3)! = (exp(2x)-2*sin(x)*exp(x)-1)/(2*exp(x)+cos(x)*(exp(2x)+1)). It appears that a(n) = (4n+3)!*coefficient of x^(4n+3) in the Taylor expansion of -4/(2*exp(x)+cos(x)*(exp(2x)+1)).