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A065602
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Triangle T(n,k) giving number of hill-free Dyck paths of length 2n and having height of first peak equal to k.
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4
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1, 1, 1, 3, 2, 1, 8, 6, 3, 1, 24, 18, 10, 4, 1, 75, 57, 33, 15, 5, 1, 243, 186, 111, 54, 21, 6, 1, 808, 622, 379, 193, 82, 28, 7, 1, 2742, 2120, 1312, 690, 311, 118, 36, 8, 1, 9458, 7338, 4596, 2476, 1164, 474, 163, 45, 9, 1, 33062, 25724, 16266, 8928, 4332, 1856, 692, 218, 55, 10, 1
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OFFSET
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2,4
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COMMENTS
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A Riordan triangle.
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LINKS
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FORMULA
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Sum_{k=2..n} T(n, k) = A000957(n+1).
T(n, k) = Sum_{j=0..floor((n-k)/2)} (k-1+2*j)*binomial(2*n-k-1-2*j, n-1)/(2*n-k-1-2*j).
G.f.: t^2*z^2*C/( (1-z^2*C^2)*(1-t*z*C) ), where C = (1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
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EXAMPLE
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T(3,2)=1 reflecting the unique Dyck path (UUDUDD) of length 6, with no hills and height of first peak equal to 2.
Triangle begins:
1;
1, 1;
3, 2, 1;
8, 6, 3, 1;
24, 18, 10, 4, 1;
75, 57, 33, 15, 5, 1;
243, 186, 111, 54, 21, 6, 1;
808, 622, 379, 193, 82, 28, 7, 1;
2742, 2120, 1312, 690, 311, 118, 36, 8, 1;
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MAPLE
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a := proc(n, k) if n=0 and k=0 then 1 elif k<2 or k>n then 0 else sum((k-1+2*j)*binomial(2*n-k-1-2*j, n-1)/(2*n-k-1-2*j), j=0..floor((n-k)/2)) fi end: seq(seq(a(n, k), k=2..n), n=1..14);
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MATHEMATICA
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nmax = 12; t[n_, k_] := Sum[(k-1+2j)*Binomial[2n-k-1-2j, n-1] / (2n-k-1-2j), {j, 0, (n-k)/2}]; Flatten[ Table[t[n, k], {n, 2, nmax}, {k, 2, n}]] (* Jean-François Alcover, Nov 08 2011, after Maple *)
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PROG
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(Haskell)
a065602 n k = sum
[(k-1+2*j) * a007318' (2*n-k-1-2*j) (n-1) `div` (2*n-k-1-2*j) |
j <- [0 .. div (n-k) 2]]
a065602_row n = map (a065602 n) [2..n]
a065602_tabl = map a065602_row [2..]
(SageMath)
def T(n, k): return sum( (k+2*j-1)*binomial(2*n-2*j-k-1, n-1)/(2*n-2*j-k-1) for j in (0..(n-k)//2) )
flatten([[T(n, k) for k in (2..n)] for n in (2..12)]) # G. C. Greubel, May 26 2022
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CROSSREFS
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Row sums give A000957 (the Fine sequence).
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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