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A147309
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Riordan array [sec(x), log(sec(x) + tan(x))].
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14
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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, 0, 560, 0, 56, 0, 1, 1385, 0, 4996, 0, 1470, 0, 84, 0, 1, 0, 24320, 0, 22720, 0, 3360, 0, 120, 0, 1
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OFFSET
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0,8
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COMMENTS
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Production array is [cosh(x),x] beheaded. Inverse is A147308. Row sums are A000111(n+1).
Define a polynomial sequence {Z(n,x)} n >= 0 by means of the recursion
(1)... Z(n+1,x) = 1/2*x*{Z(n,x-1)+Z(n,x+1)}
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
... Z(1,x) = x
... Z(2,x) = x^2
... Z(3,x) = x + x^3
... Z(4,x) = 4*x^2 + x^4
... Z(5,x) = 5*x + 10*x^3 + x^5.
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.
{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:
(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],
where {B_k} k >= 0 = [1, -1/2, 1/6, 0, -1/30, ...] is the sequence of Bernoulli numbers.
(End)
[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
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
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LINKS
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FORMULA
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GENERATING FUNCTION
The e.g.f., upon including a constant term of '1', is given by:
(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! + ....
Other forms include
(2) F(x,t) = exp(x*arcsinh(tan(t))) = exp(2*x*arctanh(tan(t/2))).
(3) F(x,t) = exp(x*(t + t^3/3! + 5*t^5/5! + 61*t^7/7! + ...)),
where the coefficients [1,1,5,61,...] are the secant or zig numbers A000364.
ROW GENERATING POLYNOMIALS
One easily checks from (1) that
d/dt(F(x,t)) = 1/2*x*(F(x-1,t) + F(x+1,t))
and so the row generating polynomials Z(n,x) satisfy the recurrence relation
(4) Z(n+1,x) = 1/2*x*{Z(n,x-1) + Z(n,x+1)}.
The e.g.f. for the odd-indexed row polynomials is
(5) sinh(x*arcsinh(tan(t))) = Sum_{n>=0} Z(2n+1,x)*t^(2n+1)/(2n+1)!.
The e.g.f. for the even-indexed row polynomials is
(6) cosh(x*arcsinh(tan(t))) = Sum_{n>=0} Z(2n,x)*t^(2n)/(2n)!.
From sinh(2*x) = 2*sinh(x)*cosh(x) we obtain the identity
(7) Z(2n+1,2*x) = 2*Sum_{k=0..n} binomial(2n+1,2k)*Z(2k,x)*Z(2n-2k+1,x).
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)).
BINOMIAL EXPANSION
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
(8) Z(n,x+y) = Sum_{k=0..n} binomial(n,k)*Z(k,x)*Z(n-k,y).
The delta operator D* associated with this binomial type sequence is
(9) D* = D - D^3/3! + 5*D^5/5! - 61*D^7/7! + 1385*D^9/9! - ..., and satisfies
the relation
(10) tan(D*)+sec(D*) = exp(D).
The delta operator D* acts as a lowering operator on the zigzag polynomials:
(11) (D*)Z(n,x) = n*Z(n-1,x).
ANALOG OF THE LITTLE FERMAT THEOREM
For integer x and odd prime p
(12) Z(p,x) = (-1)^((p-1)/2)*x (mod p).
More generally, for k = 1,2,3,...
(13) Z(p+k-1,x) = (-1)^((p-1)/2)*Z(k,x) (mod p).
RELATIONS WITH OTHER SEQUENCES
Row sums [1,1,2,5,16,61,...] are the zigzag numbers A000111(n) for n >= 1.
Column 1 (with 0's omitted) is the sequence of Euler numbers A000364.
A145876(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n+1,k-j)*Z(n,j).
A147315(n-1,k-1) = (1/k!)*Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j).
A185421(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j).
A012123(n) = (-i)^n*Z(n,i) where i = sqrt(-1). A012259(n) = 2^n*Z(n,1/2).
(End)
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
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EXAMPLE
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Triangle begins
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;
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MAPLE
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Z := proc(n, x) option remember;
description 'zigzag polynomials Z(n, x)'
if n = 0 return 1 else return 1/2*x*(Z(n-1, x-1)+Z(n-1, x+1)) end proc:
with(PolynomialTools):
for n from 1 to 10 CoefficientList(Z(n, x), x); end do; # Peter Bala, Jan 26 2011
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MATHEMATICA
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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 *)
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PROG
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(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
(Sage)
R = PolynomialRing(QQ, 'x')
@CachedFunction
def zzp(n, x) :
return 1 if n == 0 else x*(zzp(n-1, x-1)+zzp(n-1, x+1))/2
x = R.gen()
L = list(R(zzp(n, x)))
del L[0]
return L
(Sage) # uses[bell_matrix from A264428]
# Alternative: Adds a column 1, 0, 0, 0, ... at the left side of the triangle.
bell_matrix(lambda n: abs(euler_number(n)), 10) # Peter Luschny, Jan 18 2016
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CROSSREFS
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Cf. A000111 (row sums), A000364, A012123, A012259, A145876, A147315, A185415, A185417, A185419, A185421, A185424. - Peter Bala, Jan 26 2011
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KEYWORD
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AUTHOR
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STATUS
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approved
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