cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A114593 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).

Original entry on oeis.org

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, 28028, 429
Offset: 2

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Author

Emeric Deutsch, Dec 11 2005

Keywords

Comments

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=1..floor(n/2)} k*T(n,k) yields A114594.
T(n,k) is the number of permutations pi of [n-1] with k valleys such that s(pi) avoids the patterns 132, 231, and 312, where s is West's stack-sorting map. - Colin Defant, Sep 16 2018

Examples

			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;
		

Crossrefs

Programs

  • Maple
    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;
  • Mathematica
    m = 13(*rows*); G = 0; Do[G = Series[(1 + G^2 (2 + t z) z)/(1 + 2 z), {t, 0, m+1}, {z, 0, m+1}] // Normal // Expand, {m+2}]; Rest[CoefficientList[ #, t]]& /@ CoefficientList[G-1, z][[3;;]] // Flatten (* Jean-François Alcover, Jan 22 2019 *)

Formula

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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