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A104035
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Triangle T(n,k), 0 <= k <= n, read by rows, defined by T(0,0) = 1; T(0,k) = 0 if k>0 or if k<0; T(n,k) = k*T(n-1,k-1) + (k+1)*T(n-1,k+1).
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19
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1, 0, 1, 1, 0, 2, 0, 5, 0, 6, 5, 0, 28, 0, 24, 0, 61, 0, 180, 0, 120, 61, 0, 662, 0, 1320, 0, 720, 0, 1385, 0, 7266, 0, 10920, 0, 5040, 1385, 0, 24568, 0, 83664, 0, 100800, 0, 40320, 0, 50521, 0, 408360, 0, 1023120, 0, 1028160, 0, 362880, 50521, 0, 1326122, 0, 6749040
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OFFSET
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0,6
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COMMENTS
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Or, triangle of coefficients (with exponents in increasing order) in polynomials Q_n(u) defined by d^n sec x / dx^n = Q_n(tan x)*sec x.
Interpolates between factorials and Euler (or secant) numbers. Related to Springer numbers.
Companion triangles are A155100 (derivative polynomials of tangent function) and A185896 (derivative polynomials of squared secant function).
A combinatorial interpretation for the polynomial Q_n(u) as the generating function for a sign change statistic on certain types of signed permutation can be found in [Verges]. A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...,|x_n|} = {1,2,...,n}. They form a group, the hyperoctahedral group of order 2^n*n! = A000165(n), isomorphic to the group of symmetries of the n dimensional cube.
Let x_1,...,x_n be a signed permutation. Adjoin x_0 = 0 to the front of the permutation and x_(n+1) = (-1)^n*(n+1) to the end to form x_0,x_1,...,x_n,x_(n+1). Then x_0,x_1,...,x_n,x_(n+1) is a snake of type S(n;0) when x_0 < x_1 > x_2 < ... x_(n+1). For example, 0 3 -1 2 -4 is a snake of type S(3;0).
Let sc be the number of sign changes through a snake ... sc = #{i, 0 <= i <= n, x_i*x_(i+1) < 0}. For example, the snake 0 3 -1 2 -4 has sc = 3. The polynomial Q_n(u) is the generating function for the sign change statistic on snakes of type S(n;0): ... Q_n(u) = sum {snakes in S(n;0)} u^sc. See the example section below for the cases n = 2 and n = 3.
PRODUCTION MATRIX
Let D = subdiag(1,2,3,...) be the array with the indicated sequence on the first subdiagonal and zeros elsewhere and let C = transpose(D). The production matrix for this triangle is C+D: the first row of (C+D)^n is the n-th row of this triangle. D represents the derivative operator d/dx and C represents the operator p(x) -> x*d/dx(x*p(x)) acting on the basis monomials {x^n}n>=0. See Formula (1) below.
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REFERENCES
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R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998, p. 287.
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see pp. 445 and 469.
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LINKS
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FORMULA
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T(n, n) = n!; T(n, 0) = 0 if n = 2m + 1; T(n, 0) = A000364(m) if n = 2m.
Sum_{k>=0} T(m, k)*T(n, k) = T(m+n, 0).
Sum_{k>=0} T(n, k) = A001586(n): Springer numbers.
G.f.: Sum_{n >= 0} Q_n(u)*t^n/n! = 1/(cos t - u sin t).
RECURRENCE RELATION
For n>=0,
(1)... Q_(n+1)(u) = d/du Q_n(u) + u*d/du(u*Q_n(u))
... = (1+u^2)*d/du Q_n(u) + u*Q_n(u),
with starting condition Q_0(u) = 1. Compare with Formula (4) of A186492.
RELATION WITH TYPE B EULERIAN NUMBERS
(2)... Q_n(u) = ((u+i)/2)^n*B(n,(u-i)/(u+i)), where i = sqrt(-1) and
[B(n,u)]n>=0 = [1,1+u,1+6*u+u^2,1+23*u+23*u^2+u^3,...] is the sequence of type B Eulerian polynomials (with a factor of u removed) - see A060187.
(End)
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EXAMPLE
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The polynomials Q_0(u) through Q_6(u) (with exponents in decreasing order) are:
1
u
2*u^2 + 1
6*u^3 + 5*u
24*u^4 + 28*u^2 + 5
120*u^5 + 180*u^3 + 61*u
720*u^6 + 1320*u^4 + 662*u^2 + 61
Triangle begins:
1
0 1
1 0 2
0 5 0 6
5 0 28 0 24
0 61 0 180 0 120
61 0 662 0 1320 0 720
0 1385 0 7266 0 10920 0 5040
1385 0 24568 0 83664 0 100800 0 40320
0 50521 0 408360 0 1023120 0 1028160 0 362880
50521 0 1326122 0 6749040 0 13335840 0 11491200 0 3628800
0 2702765 0 30974526 0 113760240 0 185280480 0 139708800 0 39916800
2702765 0 98329108 0 692699304 0 1979524800 0 2739623040 0 1836172800 0 479001600
Examples of sign change statistic sc on snakes of type S(n;0)
= = = = = = = = = = = = = = = = = = = = = =
.....Snakes....# sign changes sc.......u^sc
= = = = = = = = = = = = = = = = = = = = = =
n=2
...0 1 -2 3...........2.................u^2
...0 2 1 3...........0.................1
...0 2 -1 3...........2.................u^2
yields Q_2(u) = 2*u^2 + 1.
n=3
...0 1 -2 3 -4.......3.................u^3
...0 1 -3 2 -4.......3.................u^3
...0 1 -3 -2 -4.......1.................u
...0 2 1 3 -4.......1.................u
...0 2 -1 3 -4.......3.................u^3
...0 2 -3 1 -4.......3.................u^3
...0 2 -3 -2 -4.......1.................u
...0 3 1 2 -4.......1.................u
...0 3 -1 2 -4.......3.................u^3
...0 3 -2 1 -4.......3.................u^3
...0 3 -2 -1 -4.......1.................u
yields Q_3(u) = 6*u^3 + 5*u.
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MATHEMATICA
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nmax = 10; t[n_, k_] := t[n, k] = k*t[n-1, k-1] + (k+1)*t[n-1, k+1]; t[0, 0] = 1; t[0, _] = 0; Flatten[ Table[t[n, k], {n, 0, nmax}, {k, 0, n}]] (* Jean-François Alcover, Nov 14 2011 *)
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PROG
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(Haskell)
a104035 n k = a104035_tabl !! n !! k
a104035_row n = a104035_tabl !! n
a104035_tabl = iterate f [1] where
f xs = zipWith (+)
(zipWith (*) [1..] (tail xs) ++ [0, 0]) ([0] ++ zipWith (*) [1..] xs)
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CROSSREFS
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See A008294 for another version of this triangle.
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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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