A104035 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).

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

Offset: 0

Author

Philippe Deléham, Apr 06 2005

Keywords

• nonn
• easy
• tabl
• nice

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.

Examples

			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.
		

References

• 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.

Links

• Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
• K. Boyadzhiev, Derivative Polynomials for tanh, tan, sech and sec in Explicit Form, arXiv:0903.0117 [math.CA], 2009-2010.
• M.-P. Grosset and A. P. Veselov, Bernoulli numbers and solitons, arXiv:math/0503175 [math.GM], 2005.
• Gordon Haigh, 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.
• Michael E. Hoffman, Derivative Polynomials, Euler Polynomials, and Associated Integer Sequences, The Electronic Journal of Combinatorics, Volume 6.1 (1999): Research paper R21, 13 p.
• M. Josuat-Vergès, Enumeration of snakes and cycle-alternating permutations, arXiv:1011.0929 [math.CO], 2010.
• Donald E. Knuth and Thomas J. Buckholtz, Computation of tangent, Euler and Bernoulli numbers, Math. Comp. 21 1967 663-688.

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.

Cf. A060187, A155100, A185896, A186492.

Programs

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)
-- Reinhard Zumkeller, Apr 27 2014

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}]] (* _Jean-François Alcover, Nov 14 2011 *)

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: (Start)

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)

T(n,0) = abs(A122045(n)). - Reinhard Zumkeller, Apr 27 2014

Extensions

Entry revised by N. J. A. Sloane, Nov 06 2009