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Search: id:A114593
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| A114593 |
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Triangle read by rows: T(n,k) is the number of hill-free Dyck paths of semilength n, having k ascents of length at least 2 (1<=k<=floor(n/2), n>=2). |
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+0
1
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| 1, 2, 4, 2, 8, 10, 16, 36, 5, 32, 112, 42, 64, 320, 224, 14, 128, 864, 960, 168, 256, 2240, 3600, 1200, 42, 512, 5632, 12320, 6600, 660, 1024, 13824, 39424, 30800, 5940, 132, 2048, 33280, 119808, 128128, 40040, 2574, 4096, 78848, 349440, 489216, 224224
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OFFSET
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2,2
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COMMENT
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Row n has floor(n/2) terms. Row sums are the Fine numbers (A000957). T(n,1)=2^(n-2). T(n,2)=n(n-3)2^(n-5) (n>4) (2*A001793). T(2n,n)=Catalan(n). T(2n+1,n)=n*Catalan(n+1). Sum(k*T(n,k),k=1..floor(n/2)) yields A114594.
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FORMULA
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T(n, k)=2^(n-2k)*binomial(n+1, k)binomial(n-k-1, k-1)/(n+1) (1<=k<=floor(n/2)). G.f.=G-1, where G=G(t, z) satisfies z(2+tz)G^2-(1+2z)G+1=0.
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EXAMPLE
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T(4,2)=2 because we have (UU)D(UU)DDD and (UU)DD(UU)DD, where U=(1,1), D=(1,-1) (ascents of length at least two are shown between parentheses).
Triangle starts:
1;
2;
4,2;
8,10;
16,36,5;
32,112,42;
64,320,224,14;
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MAPLE
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T:=proc(n, k) if k<=floor(n/2) then 2^(n-2*k)*binomial(n+1, k)*binomial(n-k-1, k-1)/(n+1) else 0 fi end: for n from 2 to 14 do seq(T(n, k), k=1..floor(n/2)) od;
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CROSSREFS
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Cf. A000957, A001793, A114594.
Sequence in context: A152874 A065286 A068217 this_sequence A114655 A051288 A120434
Adjacent sequences: A114590 A114591 A114592 this_sequence A114594 A114595 A114596
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Dec 11 2005
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