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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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4
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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; refs; listen; history; internal format)
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OFFSET
| 1,4
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COMMENTS
| 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).
From Peter Bala, June 11 2011: (Start)
Koutras has defined generalized Eulerian numbers associated with a sequence of polynomials - the ordinary Eulerian numbers A008292 being associated with the sequence of monomials x^n. The present array is the triangle of Eulerian numbers associated with the sequence of zigzag polynomials Z(n,x) defined in A147309.
See A109449, A147315 and A185424 for the respective analogs of the Pascal triangle, the Stirling numbers of the second kind and the Bernoulli numbers, associated with the sequence of zigzag polynomials.
(End)
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REFERENCES
| D. Chebikin, Variations on descents and inversions in permutations, The Electronic J. of Combinatorics, 15 (2008), #R132.
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LINKS
| M. V. Koutras, Eulerian numbers associated with sequences of polynomials, The Fibonacci Quarterly, 32 (1994), 44-57.
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FORMULA
| 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).
From Peter Bala, June 11 2011: (Start)
T(n,k) = sum {j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*Z(n,j), where Z(n,x) are the zigzag polynomials defined in A147309.
Let M denote the triangular array A109449. The first column of the array (I-x*M)^-1 is a sequence of rational functions in x whose numerator polynomials are the row polynomials of the present array.
(End)
From Vladimir Shevelev, Jul 01 2011: (Start)
a(2^(2*n-1)-2^(n-1)+1) == 1 (mod 2^n).
If n is odd prime, then a(2*n^2-n+1) == 1 (mod 2*n) and a((n^2-n+2)/2) == (-1)^((n-1)/2).
(End)
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EXAMPLE
| 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
| 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
| A000142, A000111, A001710, A008292, A109449, A147315, A185424
Sequence in context: A097006 A033306 A136347 * A039878 A162145 A039886
Adjacent sequences: A145873 A145874 A145875 * A145877 A145878 A145879
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KEYWORD
| nonn,tabl
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Oct 22 2008
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