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A132887 Number of symmetric paths in the first quadrant, from (0,0) to (n,0), consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0). 0
1, 1, 3, 2, 8, 6, 23, 17, 68, 51, 205, 154, 627, 473, 1937, 1464, 6032, 4568, 18900, 14332, 59519, 45187, 188211, 143024, 597241, 454217, 1900821, 1446604, 6065180, 4618576, 19396027 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

a(2n+1)=A059398(n); a(2n)=A059398(n-1)+A059398(n). The number of all paths in the first quadrant, from (0,0) to (n,0), consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0) is A128720(n).

LINKS

Table of n, a(n) for n=0..30.

FORMULA

G.f.=2(1+z+z^2)/[1-3z^2-z^4+sqrt((1+z^2-z^4)(1-3z^2-z^4))].

Conjecture: (n+2)*a(n) +n*a(n-1) +(-n-6)*a(n-2) -2*n*a(n-3) +7*(-n+2)*a(n-4) +5*(-n+4)*a(n-5) +3*(-n+6)*a(n-6) +2*(n-8)*a(n-7) +(3*n-26)*a(n-8) +(n-8)*a(n-9) +(n-10)*a(n-10)=0. - R. J. Mathar, Oct 08 2016

EXAMPLE

a(4)=8 because we have hhhh, hHh, HH, hUDh, UDUD, UhhD, UHD and UUDD.

MAPLE

G:=(2*(1+z+z^2))/(1-3*z^2-z^4+sqrt((1+z^2-z^4)*(1-3*z^2-z^4))): Gser:=series(G, z=0, 35): seq(coeff(Gser, z, n), n=0..30);

CROSSREFS

Cf. A128720, A059398.

Sequence in context: A122297 A073283 A117822 * A092174 A245781 A239803

Adjacent sequences:  A132884 A132885 A132886 * A132888 A132889 A132890

KEYWORD

nonn

AUTHOR

Emeric Deutsch, Sep 05 2007

STATUS

approved

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Last modified February 26 18:29 EST 2020. Contains 332293 sequences. (Running on oeis4.)