|
|
A071356
|
|
Expansion of (1 - 2*x - sqrt(1 - 4*x - 4*x^2))/(4*x^2).
|
|
21
|
|
|
1, 2, 6, 20, 72, 272, 1064, 4272, 17504, 72896, 307648, 1312896, 5655808, 24562176, 107419264, 472675072, 2091206144, 9296612352, 41507566592, 186045061120, 836830457856, 3776131489792, 17089399689216, 77548125675520, 352766964908032
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
Number of underdiagonal lattice paths from (0,0) to the line x=n, using only steps R=(1,0), V=(0,1) and D=(1,2). Also number of Motzkin paths of length n in which both the "up" and the "level" steps come in two colors. E.g., a(2)=6 because we have RR, RVR, RRV, RD, RVRV and RRVV. - Emeric Deutsch, Dec 21 2003
Inverse binomial transform of little Schroeder numbers 1,3,11,... (A001003 with first term deleted). - David Callan, Feb 07 2004
a(n) is the number of planar trees satisfying: 1) Every internal node has at least two children, 2) Among the children of a node, only the leftmost and the rightmost children can be leaves, 3) The tree has n+1 leaves. For instance, a(3)=6. - Marcelo Aguiar (maguiar(AT)math.tamu.edu), Oct 14 2005
Conjecturally, the number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) > e(j) <= e(k). [Martinez and Savage, 2.19] - Eric M. Schmidt, Jul 17 2017
Let s denote West's stack-sorting map, and let Av_n(tau_1, ..., tau_r) denote the set of permutations of [n] that avoid the patterns tau_1, ..., tau_r. It is conjectured that a(n) = |s^{-1}(Av_{n+1}(132, 231))| = |s^{-1}(Av_{n+1}(132, 312))| = |s^{-1}(Av_{n+1}(231, 312))|. Only the last of these equalities is known. - Colin Defant, Sep 16 2018
|
|
REFERENCES
|
Bin Han, The gamma-positive coefficients arising in segmented permutations, Discrete Math., 344 (2012), #112336. See p. 7.
|
|
LINKS
|
G. Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973), p. 32-33 (same sequence but with offset 1).
G. Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973). (Annotated scanned copy)
|
|
FORMULA
|
G.f. A(x) satisfies 2x^2*A(x)^2+(2x-1)*A(x)+1=0 and A(x)=1/(1-2x-2x^2/A(x)). - Michael Somos, Sep 06 2003
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k)C(k)2^(n-2k)*2^k. - Paul Barry, May 18 2005
G.f.: (1 - 2*x - sqrt(1 - 4*x - 4*x^2) )/(4*x^2) = 2/(1 - 2*x +sqrt(1 - 4*x - 4*x^2)).
Moment representation is a(n) = (1/(4*Pi))*int(x^n*sqrt(4-4x-x^2), x, -2*sqrt(2)-2, 2*sqrt(2)-2). - Paul Barry, Jan 08 2008
G.f.: 1/(1-2x-2x^2/(1-2x-2x^2/(1-2x-2x^2/(1-2x-2x^2/(1-2x-2x^2/.... (continued fraction). - Paul Barry, Dec 06 2008
a(n) = sum of top row terms of M^n, M = an infinite square production matrix as follows:
1, 1, 0, 0, 0, 0, ...
2, 1, 1, 0, 0, 0, ...
2, 2, 1, 1, 0, 0, ...
2, 2, 2, 1, 1, 0, ...
2, 2, 2, 2, 1, 1, ...
2, 2, 2, 2, 2, 1, ... (End)
E.g.f.: a(n) = n!* [x^n] exp(2*x)*BesselI(1, 2*sqrt(2)*x)/(sqrt(2)*x). - Peter Luschny, Aug 25 2012
D-finite with recurrence: (n+2)*a(n) +2*(-2*n-1)*a(n-1) +4*(-n+1)*a(n-2)=0. - R. J. Mathar, Dec 02 2012 (Formula verified and used for computations. - Fung Lam, Feb 24 2014)
a(n) ~ 2^(n - 1/4) * (1+sqrt(2))^(n + 3/2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Sep 24 2013, simplified Jan 26 2019
a(n) = 2^n * hypergeom((1 - n)/2, -n/2, 2, 2). - Peter Luschny, May 30 2021
|
|
EXAMPLE
|
a(3) = 20 = sum of top row terms in M^3 = (9 + 7 + 3 + 1).
|
|
MATHEMATICA
|
CoefficientList[Series[(1-2*x-Sqrt[1-4*x-4*x^2])/(4*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 24 2013 *)
a[n_] := 2^n Hypergeometric2F1[(1-n)/2, -n/2, 2, 2];
|
|
PROG
|
(PARI) a(n)=if(n<0, 0, n++; polcoeff(serreverse(x/(1+2*x+2*x^2)+x*O(x^n)), n))
(PARI) {a(n)= if(n<1, n==0, polcoeff( 2/(1 -2*x +sqrt(1 -4*x -4*x^2 +x*O(x^n))), n))}
(PARI) {a(n)= local(A); if(n<0, 0, A= x*O(x^n); n!*simplify(polcoeff( exp(2*x +A)* besseli(1, 2*x* quadgen(8) +A), n)))} /* Michael Somos, Mar 31 2007 */
(Sage)
T = [0]*(n+1); R = [1]
for m in (1..n-1):
a, b, c = 1, 0, 0
for k in range(m, -1, -1):
r = a + 2*(b + c)
if k < m : T[k+2] = u;
a, b, c = T[k-1], a, b
u = r
T[1] = u; R.append(u)
return R
(Magma) R<x>:=PowerSeriesRing(Rationals(), 30); Coefficients(R!((1 - 2*x - Sqrt(1 - 4*x - 4*x^2))/(4*x^2))); // Vincenzo Librandi, Jan 21 2020
|
|
CROSSREFS
|
A036774(n) = a(n-1) * n! / 2^(n-1).
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|