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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
1, 1, 2, 5, 12, 30, 82, 237, 708, 2188, 6980, 22814, 75994, 257266, 883006, 3065757, 10748620, 38005844, 135385700, 485439532, 1750738084, 6347006468, 23118315044, 84565309214, 310536661002, 1144393816154, 4231119156334
OFFSET
0,3
COMMENTS
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
REFERENCES
R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999, p. 179.
FORMULA
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.
EXAMPLE
a(4)=12 because among the 14 Dyck paths of semilength 4 the only
counterexamples are UUDUUDDD and UUUDDUDD, where U=(1,1), D=(1,-1).
PROG
(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)))
CROSSREFS
Sequence in context: A182488 A363912 A261788 * A309506 A319557 A261937
KEYWORD
nonn
AUTHOR
Emeric Deutsch, Dec 08 2005
STATUS
approved