OFFSET
1,3
COMMENTS
Define a polynomial sequence S(n,x) recursively by
(1)... S(n+1,x) = x*S(n,x-1)+(x+1)*S(n,x+1) with S(0,x) = 1.
This table lists the coefficients of these polynomials (for n>=1) in ascending powers of x.
The first few polynomials are
S(0,x) = 1
S(1,x) = 2*x+1
S(2,x) = 4*x^2+4*x+3
S(3,x) = 8*x^3+12*x^2+26*x+11.
The sequence [1,1,3,11,57,...] of constant terms of the polynomials is the sequence of Springer numbers A001586. The zeros of the polynomials S(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
Compare the recurrence (1) with the recurrence relation satisfied by the coefficients T(n,k) of the polynomials of A104035, namely
(2)... T(n+1,k) = k*T(n,k-1)+(k+1)*T(n,k+1).
LINKS
FORMULA
E.g.f: F(x,t) = 1/(cos(t)-sin(t))*(tan(2*t)+sec(2*t))^x
= (cos(t)+sin(t))^x/(cos(t)-sin(t))^(x+1)
= 1 + (2*x+1)*t + (4*x^2+4*x+3)*t^2/2! + ....
Note that (tan(t)+sec(t))^x is the e.g.f for table A147309.
ROW POLYNOMIALS
The easily checked identity d/dt F(x,t) = x*F(x-1,t)+(x+1)*F(x+1,t) shows that the row generating polynomials of this table are the polynomials S(n,x) described in the Comments section above.
The polynomials S(n,-x) satisfy a Riemann hypothesis: that is, the zeros of S(n,-x) lie on the vertical line Re(x) = 1/2 in the complex plane - see the link.
RELATION WITH OTHER SEQUENCES
1st column [1,1,3,11,57,...] is A001586.
Row sums sequence [1,3,11,57,...] is also A001586.
For n>=1, the values 1/2^n*P(2*n,-1/2) = [1,7,139,5473,...] appear to be A126156.
EXAMPLE
Table begin
n\k|.....0.....1.....2.....3.....4.....5......6
===============================================
0..|.....1
1..|.....1.....2
2..|.....3.....4.....4
3..|....11....26....12.....8
4..|....57...120...136....32...16
5..|...361...970...760...560...80.....32
6..|..2763..7052..8860..3680..2000...192....64
...
MAPLE
S := proc(n, x) option remember;
description 'polynomials S(n, x)'
if n = 0 return 1 else return x*S(n-1, x-1)+(x+1)*S(n-1, x+1)
end proc:
with(PolynomialTools):
for n from 1 to 10 CoefficientList(S(n, x), x); end do;
MATHEMATICA
S[0, _] = 1; S[n_, x_] := S[n, x] = x*S[n-1, x-1] + (x+1)*S[n-1, x+1]; Table[ CoefficientList[S[n, x], x], {n, 0, 8}] // Flatten (* Jean-François Alcover, Apr 15 2015 *)
CROSSREFS
KEYWORD
AUTHOR
Peter Bala, Jan 28 2011
STATUS
approved