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%I #57 Mar 14 2020 17:08:08
%S 1,0,1,1,0,1,0,4,0,1,5,0,10,0,1,0,40,0,20,0,1,61,0,175,0,35,0,1,0,768,
%T 0,560,0,56,0,1,1385,0,4996,0,1470,0,84,0,1,0,24320,0,22720,0,3360,0,
%U 120,0,1
%N Riordan array [sec(x), log(sec(x) + tan(x))].
%C Production array is [cosh(x),x] beheaded. Inverse is A147308. Row sums are A000111(n+1).
%C Unsigned version of A147308. - _N. J. A. Sloane_, Nov 07 2008
%C From _Peter Bala_, Jan 26 2011: (Start)
%C Define a polynomial sequence {Z(n,x)} n >= 0 by means of the recursion
%C (1)... Z(n+1,x) = 1/2*x*{Z(n,x-1)+Z(n,x+1)}
%C with starting condition Z(0,x) = 1. We call Z(n,x) the zigzag polynomial of degree n. This table lists the coefficients of these polynomials (for n >= 1) in ascending powers of x, row indices shifted by 1. The first few polynomials are
%C ... Z(1,x) = x
%C ... Z(2,x) = x^2
%C ... Z(3,x) = x + x^3
%C ... Z(4,x) = 4*x^2 + x^4
%C ... Z(5,x) = 5*x + 10*x^3 + x^5.
%C The value Z(n,1) equals the zigzag number A000111(n). The polynomials Z(n,x) occur in formulas for the enumeration of permutations by alternating descents A145876 and in the enumeration of forests of non-plane unary binary labeled trees A147315.
%C {Z(n,x)}n>=0 is a polynomial sequence of binomial type and so is analogous to the sequence of monomials x^n. Denoting Z(n,x) by x^[n] to emphasize this analogy, we have, for example, the following analog of Bernoulli's formula for the sum of integer powers:
%C (2)... 1^[m]+...+(n-1)^[m] = (1/(m+1))*Sum_{k=0..m} (-1)^floor(k/2)*binomial(m+1,k)*B_k*n^[m+1-k],
%C where {B_k} k >= 0 = [1, -1/2, 1/6, 0, -1/30, ...] is the sequence of Bernoulli numbers.
%C For similarly defined polynomial sequences to Z(n,x) see A185415, A185417 and A185419. See also A185424.
%C (End)
%C [gd(x)^(-1)]^m = Sum_{n>=m} Tg(n,m)*(m!/n!)*x^n, where gd(x) is Gudermannian function, Tg(n+1,m+1)=T(n,m). - _Vladimir Kruchinin_, Dec 18 2011
%C The Bell transform of abs(E(n)), E(n) the Euler numbers. For the definition of the Bell transform see A264428. - _Peter Luschny_, Jan 18 2016
%H G. C. Greubel, <a href="/A147309/b147309.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%F From _Peter Bala_, Jan 26 2011: (Start)
%F GENERATING FUNCTION
%F The e.g.f., upon including a constant term of '1', is given by:
%F (1) F(x,t) = (tan(t) + sec(t))^x = Sum_{n>=0} Z(n,x)*t^n/n! = 1 + x*t + x^2*t^2/2! + (x+x^3)*t^3/3! + ....
%F Other forms include
%F (2) F(x,t) = exp(x*arcsinh(tan(t))) = exp(2*x*arctanh(tan(t/2))).
%F (3) F(x,t) = exp(x*(t + t^3/3! + 5*t^5/5! + 61*t^7/7! + ...)),
%F where the coefficients [1,1,5,61,...] are the secant or zig numbers A000364.
%F ROW GENERATING POLYNOMIALS
%F One easily checks from (1) that
%F d/dt(F(x,t)) = 1/2*x*(F(x-1,t) + F(x+1,t))
%F and so the row generating polynomials Z(n,x) satisfy the recurrence relation
%F (4) Z(n+1,x) = 1/2*x*{Z(n,x-1) + Z(n,x+1)}.
%F The e.g.f. for the odd-indexed row polynomials is
%F (5) sinh(x*arcsinh(tan(t))) = Sum_{n>=0} Z(2n+1,x)*t^(2n+1)/(2n+1)!.
%F The e.g.f. for the even-indexed row polynomials is
%F (6) cosh(x*arcsinh(tan(t))) = Sum_{n>=0} Z(2n,x)*t^(2n)/(2n)!.
%F From sinh(2*x) = 2*sinh(x)*cosh(x) we obtain the identity
%F (7) Z(2n+1,2*x) = 2*Sum_{k=0..n} binomial(2n+1,2k)*Z(2k,x)*Z(2n-2k+1,x).
%F The zeros of Z(n,x) lie on the imaginary axis (use (4) and adapt the proof given in A185417 for the zeros of the polynomial S(n,x)).
%F BINOMIAL EXPANSION
%F The form of the e.g.f. shows that {Z(n,x)} n >= 0 is a sequence of polynomials of binomial type. In particular, we have the expansion
%F (8) Z(n,x+y) = Sum_{k=0..n} binomial(n,k)*Z(k,x)*Z(n-k,y).
%F The delta operator D* associated with this binomial type sequence is
%F (9) D* = D - D^3/3! + 5*D^5/5! - 61*D^7/7! + 1385*D^9/9! - ..., and satisfies
%F the relation
%F (10) tan(D*)+sec(D*) = exp(D).
%F The delta operator D* acts as a lowering operator on the zigzag polynomials:
%F (11) (D*)Z(n,x) = n*Z(n-1,x).
%F ANALOG OF THE LITTLE FERMAT THEOREM
%F For integer x and odd prime p
%F (12) Z(p,x) = (-1)^((p-1)/2)*x (mod p).
%F More generally, for k = 1,2,3,...
%F (13) Z(p+k-1,x) = (-1)^((p-1)/2)*Z(k,x) (mod p).
%F RELATIONS WITH OTHER SEQUENCES
%F Row sums [1,1,2,5,16,61,...] are the zigzag numbers A000111(n) for n >= 1.
%F Column 1 (with 0's omitted) is the sequence of Euler numbers A000364.
%F A145876(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n+1,k-j)*Z(n,j).
%F A147315(n-1,k-1) = (1/k!)*Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j).
%F A185421(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j).
%F A012123(n) = (-i)^n*Z(n,i) where i = sqrt(-1). A012259(n) = 2^n*Z(n,1/2).
%F (End)
%F T(n,m) = Sum(i=0..n-m, s(i)/(n-i)!*Sum(k=m..n-i, A147315(n-i,k)*Stirling1(k,m))), m>0, T(n,0) = s(n), s(n)=[1,0,1,0,5,0,61,0,1385,0,50521,...] (see A000364). - _Vladimir Kruchinin_, Mar 10 2011
%e Triangle begins
%e 1;
%e 0, 1;
%e 1, 0, 1;
%e 0, 4, 0, 1;
%e 5, 0, 10, 0, 1;
%e 0, 40, 0, 20, 0, 1;
%e 61, 0, 175, 0, 35, 0, 1;
%p Z := proc(n, x) option remember;
%p description 'zigzag polynomials Z(n, x)'
%p if n = 0 return 1 else return 1/2*x*(Z(n-1, x-1)+Z(n-1, x+1)) end proc:
%p with(PolynomialTools):
%p for n from 1 to 10 CoefficientList(Z(n, x), x); end do; # _Peter Bala_, Jan 26 2011
%t t[n_, k_] := SeriesCoefficient[ 2^k*ArcTan[(E^x - 1)/(E^x + 1)]^k*n!/k!, {x, 0, n}]; Table[t[n, k], {n, 1, 10}, {k, 1, n}] // Flatten // Abs (* _Jean-François Alcover_, Jan 23 2015 *)
%o (PARI) T(n, k)=local(X); if(k<1 || k>n, 0, X=x+x*O(x^n); n!*polcoeff(polcoeff((tan(X)+1/cos(X))^y, n), k)) \\ _Paul D. Hanna_, Feb 06 2011
%o (Sage)
%o R = PolynomialRing(QQ, 'x')
%o @CachedFunction
%o def zzp(n, x) :
%o return 1 if n == 0 else x*(zzp(n-1, x-1)+zzp(n-1, x+1))/2
%o def A147309_row(n) :
%o x = R.gen()
%o L = list(R(zzp(n, x)))
%o del L[0]
%o return L
%o for n in (1..10) : print(A147309_row(n)) # _Peter Luschny_, Jul 22 2012
%o (Sage) # uses[bell_matrix from A264428]
%o # Alternative: Adds a column 1,0,0,0, ... at the left side of the triangle.
%o bell_matrix(lambda n: abs(euler_number(n)), 10) # _Peter Luschny_, Jan 18 2016
%Y Cf. A000111 (row sums), A000364, A012123, A012259, A145876, A147315, A185415, A185417, A185419, A185421, A185424. - _Peter Bala_, Jan 26 2011
%K nonn,tabl
%O 0,8
%A _Paul Barry_, Nov 05 2008
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