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%I #54 Feb 07 2022 03:42:11
%S 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,
%T 720,0,1385,0,7266,0,10920,0,5040,1385,0,24568,0,83664,0,100800,0,
%U 40320,0,50521,0,408360,0,1023120,0,1028160,0,362880,50521,0,1326122,0,6749040
%N 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).
%C 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.
%C Interpolates between factorials and Euler (or secant) numbers. Related to Springer numbers.
%C Companion triangles are A155100 (derivative polynomials of tangent function) and A185896 (derivative polynomials of squared secant function).
%C 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.
%C 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).
%C 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.
%C PRODUCTION MATRIX
%C 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.
%D R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998, p. 287.
%D S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see pp. 445 and 469.
%H Reinhard Zumkeller, <a href="/A104035/b104035.txt">Rows n = 0..125 of triangle, flattened</a>
%H K. Boyadzhiev, <a href="http://arxiv.org/abs/0903.0117">Derivative Polynomials for tanh, tan, sech and sec in Explicit Form</a>, arXiv:0903.0117 [math.CA], 2009-2010.
%H M.-P. Grosset and A. P. Veselov, <a href="http://arXiv.org/abs/math.GM/0503175">Bernoulli numbers and solitons</a>, arXiv:math/0503175 [math.GM], 2005.
%H Gordon Haigh, <a href="http://www.jstor.org/stable/3615119">A "natural" approach to Pick's theorem</a>, Math. Gaz. 64 (1980), no. 429, 173-180.
%H Michael E. Hoffman, <a href="http://www.jstor.org/stable/2974853">Derivative polynomials for tangent and secant</a>, Amer. Math. Monthly, 102 (1995), 23-30.
%H Michael E. Hoffman, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v6i1r21">Derivative Polynomials, Euler Polynomials, and Associated Integer Sequences</a>, The Electronic Journal of Combinatorics, Volume 6.1 (1999): Research paper R21, 13 p.
%H M. Josuat-Vergès, <a href="http://arxiv.org/abs/1011.0929">Enumeration of snakes and cycle-alternating permutations</a>, arXiv:1011.0929 [math.CO], 2010.
%H Donald E. Knuth and Thomas J. Buckholtz, <a href="http://dx.doi.org/10.1090/S0025-5718-1967-0221735-9">Computation of tangent, Euler and Bernoulli numbers</a>, Math. Comp. 21 1967 663-688.
%F T(n, n) = n!; T(n, 0) = 0 if n = 2m + 1; T(n, 0) = A000364(m) if n = 2m.
%F Sum_{k>=0} T(m, k)*T(n, k) = T(m+n, 0).
%F Sum_{k>=0} T(n, k) = A001586(n): Springer numbers.
%F G.f.: Sum_{n >= 0} Q_n(u)*t^n/n! = 1/(cos t - u sin t).
%F From _Peter Bala_: (Start)
%F RECURRENCE RELATION
%F For n>=0,
%F (1)... Q_(n+1)(u) = d/du Q_n(u) + u*d/du(u*Q_n(u))
%F ... = (1+u^2)*d/du Q_n(u) + u*Q_n(u),
%F with starting condition Q_0(u) = 1. Compare with Formula (4) of A186492.
%F RELATION WITH TYPE B EULERIAN NUMBERS
%F (2)... Q_n(u) = ((u+i)/2)^n*B(n,(u-i)/(u+i)), where i = sqrt(-1) and
%F [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.
%F (End)
%F T(n,0) = abs(A122045(n)). - _Reinhard Zumkeller_, Apr 27 2014
%e The polynomials Q_0(u) through Q_6(u) (with exponents in decreasing order) are:
%e 1
%e u
%e 2*u^2 + 1
%e 6*u^3 + 5*u
%e 24*u^4 + 28*u^2 + 5
%e 120*u^5 + 180*u^3 + 61*u
%e 720*u^6 + 1320*u^4 + 662*u^2 + 61
%e Triangle begins:
%e 1
%e 0 1
%e 1 0 2
%e 0 5 0 6
%e 5 0 28 0 24
%e 0 61 0 180 0 120
%e 61 0 662 0 1320 0 720
%e 0 1385 0 7266 0 10920 0 5040
%e 1385 0 24568 0 83664 0 100800 0 40320
%e 0 50521 0 408360 0 1023120 0 1028160 0 362880
%e 50521 0 1326122 0 6749040 0 13335840 0 11491200 0 3628800
%e 0 2702765 0 30974526 0 113760240 0 185280480 0 139708800 0 39916800
%e 2702765 0 98329108 0 692699304 0 1979524800 0 2739623040 0 1836172800 0 479001600
%e Examples of sign change statistic sc on snakes of type S(n;0)
%e = = = = = = = = = = = = = = = = = = = = = =
%e .....Snakes....# sign changes sc.......u^sc
%e = = = = = = = = = = = = = = = = = = = = = =
%e n=2
%e ...0 1 -2 3...........2.................u^2
%e ...0 2 1 3...........0.................1
%e ...0 2 -1 3...........2.................u^2
%e yields Q_2(u) = 2*u^2 + 1.
%e n=3
%e ...0 1 -2 3 -4.......3.................u^3
%e ...0 1 -3 2 -4.......3.................u^3
%e ...0 1 -3 -2 -4.......1.................u
%e ...0 2 1 3 -4.......1.................u
%e ...0 2 -1 3 -4.......3.................u^3
%e ...0 2 -3 1 -4.......3.................u^3
%e ...0 2 -3 -2 -4.......1.................u
%e ...0 3 1 2 -4.......1.................u
%e ...0 3 -1 2 -4.......3.................u^3
%e ...0 3 -2 1 -4.......3.................u^3
%e ...0 3 -2 -1 -4.......1.................u
%e yields Q_3(u) = 6*u^3 + 5*u.
%t 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 *)
%o (Haskell)
%o a104035 n k = a104035_tabl !! n !! k
%o a104035_row n = a104035_tabl !! n
%o a104035_tabl = iterate f [1] where
%o f xs = zipWith (+)
%o (zipWith (*) [1..] (tail xs) ++ [0,0]) ([0] ++ zipWith (*) [1..] xs)
%o -- _Reinhard Zumkeller_, Apr 27 2014
%Y See A008294 for another version of this triangle.
%Y Setting u=0,1,2,3,4 gives A000364, A001586, A156129, A156131, A156132.
%Y Setting u=sqrt(2) gives A156134 and A156138; u=sqrt(3) gives A002437 and A002439.
%Y Cf. A060187, A155100, A185896, A186492.
%K nonn,easy,tabl,nice
%O 0,6
%A _Philippe Deléham_, Apr 06 2005
%E Entry revised by _N. J. A. Sloane_, Nov 06 2009
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