A345194 Number of alternating patterns of length n.
1, 1, 2, 6, 22, 102, 562, 3618, 26586, 219798, 2018686, 20393790, 224750298, 2683250082, 34498833434, 475237879950, 6983085189454, 109021986683046, 1802213242949602, 31447143854808378, 577609702827987882, 11139837273501641502, 225075546284489412854
Offset: 0
Keywords
Examples
The a(0) = 1 through a(3) = 6 alternating patterns: () (1) (1,2) (1,2,1) (2,1) (1,3,2) (2,1,2) (2,1,3) (2,3,1) (3,1,2)
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..200
Crossrefs
Programs
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Mathematica
wigQ[y_]:=Or[Length[y]==0,Length[Split[y]]== Length[y]&&Length[Split[Sign[Differences[y]]]]==Length[y]-1]; allnorm[n_]:=If[n<=0,{{}},Function[s, Array[Count[s,y_/;y<=#]+1&,n]]/@Subsets[Range[n-1]+1]]; Table[Length[Select[Join@@Permutations/@allnorm[n],wigQ]],{n,0,6}]
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PARI
F(p,x) = {sum(k=0, p, (-1)^((k+1)\2)*binomial((p+k)\2, k)*x^k)} R(n,k) = {Vec(if(k==1, x, 2*F(k-2,-x)/F(k-1,x)-2-(k-2)*x) + O(x*x^n))} seq(n)= {concat([1], sum(k=1, n, R(n, k)*sum(r=k, n, binomial(r, k)*(-1)^(r-k)) ))} \\ Andrew Howroyd, Feb 04 2022
Formula
a(n) = 2*A350354(n) for n >= 2. - Andrew Howroyd, Feb 04 2022
Extensions
a(10)-a(18) from Alois P. Heinz, Dec 10 2021
Terms a(19) and beyond from Andrew Howroyd, Feb 04 2022
Comments