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Search: id:A109187
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| A109187 |
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Triangle read by rows: T(n,k) is number of Grand Motzkin paths of length n having k (1,0)-steps. (A Grand Motzkin path is a path in the half-plane x>=0, starting at (0,0), ending at (n,0) and consisting of steps u=(1,1), d=(1,-1) and h=(1,0).). |
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+0
2
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| 1, 0, 1, 2, 0, 1, 0, 6, 0, 1, 6, 0, 12, 0, 1, 0, 30, 0, 20, 0, 1, 20, 0, 90, 0, 30, 0, 1, 0, 140, 0, 210, 0, 42, 0, 1, 70, 0, 560, 0, 420, 0, 56, 0, 1, 0, 630, 0, 1680, 0, 756, 0, 72, 0, 1, 252, 0, 3150, 0, 4200, 0, 1260, 0, 90, 0, 1, 0, 2772, 0, 11550, 0, 9240, 0, 1980, 0, 110, 0, 1, 924
(list; table; graph; listen)
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OFFSET
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0,4
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COMMENT
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Row sums yield the central trinomial coefficients (A002426). T(2n+1,0)=0; T(2n,0)=binom(2n,n) (A000984). sum(k*T(n,k),k=0..n)=A109188(n). Except for the order, same rows as those of A105868.
Coefficient array of the polynomials P(n,x)=x^n*F(1/2-n/2,-n/2;1;4/x^2). [From Paul Barry (pbarry(AT)wit.ie), Oct 04 2008]
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FORMULA
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G.f.= 1/sqrt[(1-tz)^2-4z^2].
Column k has e.g.f. (x^k/k!)*Bessel_I(0,2x). - Paul Barry (pbarry(AT)wit.ie), Mar 11 2006
T(n,k)=binomial((n+k)/2,k)*binomial(n,(n+k)/2)*(1+(-1)^(n-k))/2; - Paul Barry (pbarry(AT)wit.ie), Sep 18 2007
G.f.: 1/(1-xy-2x^2/(1-xy-x^2/(1-xy-x^2/(1-xy-x^2/(1-.... (continued fraction). [From Paul Barry (pbarry(AT)wit.ie), Jan 28 2009]
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EXAMPLE
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T(3,1)=6 because we have hud,hdu,udh,duh,uhd,dhu, where u=(1,1),d=(1,-1), h=(1,0).
Triangle begins:
1;
0,1;
2,0,1;
0,6,0,1;
6,0,12,0,1;
0,30,0,20,0,1;
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MAPLE
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G:=1/sqrt((1-t*z)^2-4*z^2):Gser:=simplify(series(G, z=0, 15)): P[0]:=1: for n from 1 to 13 do P[n]:=coeff(Gser, z^n) od: for n from 0 to 13 do seq(coeff(t*P[n], t^k), k=1..n+1) od;
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CROSSREFS
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Cf. A002426, A000984, A109188, A105868.
Sequence in context: A137526 A137525 A166335 this_sequence A166357 A067147 A112227
Adjacent sequences: A109184 A109185 A109186 this_sequence A109188 A109189 A109190
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KEYWORD
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nonn,tabl
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 21 2005
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