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A147309
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Riordan array [sec(x), log(sec(x)+tan(x))].
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13
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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
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,8
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COMMENTS
| Production array is [cosh(x),x] beheaded. Inverse is A147308. Row sums are A000111(n+1).
Unsigned version of A147308. - N. J. A. Sloane (njas(AT)research.att.com), Nov 07 2008
Contribution from Peter Bala Jan 26 2011 (Start):
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 labelled 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.
For similarly defined polynomial sequences to Z(n,x) see A185415, A185417 and A185419. See also A185424.
(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). [From Vladimir Kruchinin, Dec 18 2011]
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FORMULA
| Contribution from Peter Bala Jan 26 2011 (Start):
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..inf} 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 egf 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 egf 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 egf 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) with 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) [From Kruchinin Vladimir (kru(AT)ie.tusur.ru), Mar 10 2011]
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EXAMPLE
| 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
| 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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PROG
| (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 */
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CROSSREFS
| Cf. A000111(row sums), A000364, A012123, A012259, A145876, A147315, A185415, A185417, A185419, A185421, A185424. - Peter Bala, Jan 26 2011
Sequence in context: A186759 A065623 A178103 * A147308 A110064 A021253
Adjacent sequences: A147306 A147307 A147308 * A147310 A147311 A147312
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KEYWORD
| easy,nonn,tabl
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AUTHOR
| Paul Barry (pbarry(AT)wit.ie), Nov 05 2008
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