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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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17
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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
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,6
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COMMENTS
| 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
| R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998, p. 287.
Haigh, Gordon; A "natural" approach to Pick's theorem. Math. Gaz. 64 (1980), no. 429, 173-180.
Michael E. Hoffman, Derivative polynomials for tangent and secant, Amer. Math. Monthly, 102 (1995), 23-30.
Knuth, D. E. and Buckholtz, Thomas J., Computation of tangent, Euler and Bernoulli numbers. Math. Comp. 21 1967 663-688.
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see pp. 445 and 469.
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LINKS
| K. Boyadzhiev, Derivative Polynomials for tanh, tan, sech and sec in Explicit Form
M.-P. Grosset and A. P. Veselov, Bernoulli numbers and solitons
Michael E. Hoffman, DERIVATIVE POLYNOMIALS, EULER POLYNOMIALS, AND ASSOCIATED INTEGER SEQUENCES
M. Josiat-Verges, Enumeration of snakes and cycle-alternating permutations
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FORMULA
| 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).
From Peter Bala
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
| The polynomials Q_0(u) through Q_6(u) (with exponents in descreasing 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
| 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}]] (* From Jean-François Alcover, Nov 14 2011 *)
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CROSSREFS
| See A008294 for another version of this triangle.
Setting u=0,1,2,3,4 gives A000364, A001586, A156129, A156131, A156132.
Setting u=sqrt(2) gives A156134 and A156138; u=sqrt(3) gives A002437 and A002439. A060187, A155100, A185896, A186492.
Sequence in context: A054013 A048050 A078153 * A196409 A115333 A105523
Adjacent sequences: A104032 A104033 A104034 * A104036 A104037 A104038
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KEYWORD
| nonn,easy,tabl,nice
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AUTHOR
| Philippe DELEHAM ( kolotoko(AT)wanadoo.fr), Apr 06 2005
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EXTENSIONS
| Entry revised by N. J. A. Sloane, Nov 06 2009
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