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 A112412 Number of Dyck paths of semilength n for which the number of ascents of length 1 is equal to the number of descents of length 1. 0

%I

%S 1,1,2,5,12,30,82,237,708,2188,6980,22814,75994,257266,883006,3065757,

%T 10748620,38005844,135385700,485439532,1750738084,6347006468,

%U 23118315044,84565309214,310536661002,1144393816154,4231119156334

%N Number of Dyck paths of semilength n for which the number of ascents of length 1 is equal to the number of descents of length 1.

%C Apparently: Number of Dyck n-paths with equal numbers of peaks to the left and to the right of the midpoint (ordinate x=n). - _David Scambler_, Aug 08 2012

%D R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999, p. 179.

%F G.f. is the diagonal of g(t, s, z), where g=g(t, s, z) is defined by z(1+tz-tsz)(1+sz-tsz)g^2 - [1+(1-ts)z-(1-t)(1-s)z^2]g+1=0 (g is the trivariate g.f. of Dyck paths, where z marks semilength and t (s) marks number of ascents (descents) of length 1.

%e a(4)=12 because among the 14 Dyck paths of semilength 4 the only

%e counterexamples are UUDUUDDD and UUUDDUDD, where U=(1,1), D=(1,-1).

%o (PARI) z=x;s;t; f(g) = z*(1+t*z-t*s*z)*(1+s*z-t*s*z)*g^2-(1+(1-t*s)*z-(1-t)*(1-s)*z^2)*g+1 nxt(fx) = fx=truncate(fx);fx+=O(x^2)*x^poldegree(fx);fx+=f(fx) oo=30;g=1+O(z);for(n=1,oo,g=nxt(g)); g1=polcoeff(subst(subst(g,s,y),t,1/y),0,y); for(n=0,oo,print(n" "polcoeff(g1,n)))

%K nonn

%O 0,3

%A _Emeric Deutsch_, Dec 08 2005

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Last modified June 20 06:23 EDT 2021. Contains 345157 sequences. (Running on oeis4.)