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Search: id:A145876
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| A145876 |
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Triangle read by rows: T(n,k) is the number of permutations of [n] having k-1 alternating descents (1<=k<=n). The index i is an alternating descent of a permutation p if either i is odd and p(i)>p(i+1), or i is even and p(i)<p(i+1). |
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+0
1
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| 1, 1, 1, 2, 2, 2, 5, 7, 7, 5, 16, 26, 36, 26, 16, 61, 117, 182, 182, 117, 61, 272, 594, 1056, 1196, 1056, 594, 272, 1385, 3407, 6669, 8699, 8699, 6669, 3407, 1385, 7936, 21682, 46348, 67054, 76840, 67054, 46348, 21682, 7936, 50521, 151853, 350240, 556952
(list; table; graph; listen)
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OFFSET
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1,4
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COMMENT
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Row sums are the factorials (A000142).
T(n,1)=T(n,n)=A000111(n) (Euler or up-down numbers).
Sum(k*T(n,k),k=1..n)=(n+1)!/2=A001710(n+1).
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REFERENCES
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D. Chebikin, Variations on descents and inversions in permutations, The Electronic J. of Combinatorics, 15 (2008), #R132.
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FORMULA
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E.g.f =F(t,u)=t[1-tan(u(t-1))-sec(u(t-1))]/[tan(u(t-1))+sec(u(t-1))-t].
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EXAMPLE
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T(4,3)=7 because we have 1243, 4123, 1342, 3124, 2134, 2341 and 4321. For example, for p=1342 the alternating descent is {2,3}; indeed, 2 is even and p(2)=3<p(3)=4, while 3 is odd and p(3)=4>p(4)=2.
Triangle starts:
1;
1,1;
2,2,2;
5,7,7,5;
16,26,36,26,16;
61,117,182,182,117,61;
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MAPLE
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F:=t*(1-tan(u*(t-1))-sec(u*(t-1)))/(tan(u*(t-1))+sec(u*(t-1))-t): Fser:= simplify(series(F, u=0, 12)): for n from 0 to 10 do P[n]:=sort(expand(factorial(n)*coeff(Fser, u, n))) end do: for n to 10 do seq(coeff(P[n], t, j), j=1..n) end do; # yields sequence in triangular form
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CROSSREFS
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A000142, A000111, A001710
Sequence in context: A033306 A136347 A162145 this_sequence A039878 A039886 A109523
Adjacent sequences: A145873 A145874 A145875 this_sequence A145877 A145878 A145879
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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), Oct 22 2008
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