A323210 a(n) = 9*J(n)^2 where J(n) are the Jacobsthal numbers A001045 with J(0) = 1.

1, 9, 9, 81, 225, 1089, 3969, 16641, 65025, 263169, 1046529, 4198401, 16769025, 67125249, 268402689, 1073807361, 4294836225, 17180131329, 68718952449, 274878955521, 1099509530625, 4398050705409, 17592177655809, 70368760954881, 281474943156225, 1125899973951489

Offset: 0

Author

Peter Luschny, Jan 09 2019

Keywords

• nonn
• easy

Comments

Colin Barker conjectures that A208556 is a shifted version of this sequence.

Links

• Index entries for linear recurrences with constant coefficients, signature (3,6,-8).

Crossrefs

Cf. A001045, A062510, A139818, A208556, A322942, A323232.

Programs

Maple

gf := (8*x^3 - 24*x^2 + 6*x + 1)/((4*x - 1)*(2*x + 1)*(x - 1)):
ser := series(gf,x,32): seq(coeff(ser,x,n), n=0..25);

Mathematica

LinearRecurrence[{3, 6, -8}, {1, 9, 9, 81}, 25]

Sage

# Demonstrates the product formula.
CC = ComplexField(200)
def t(n,k): return CC(3)*cos(CC(pi*k/n)) - CC(i)*sin(CC(pi*k/n))
def T(n,k): return t(n,k)*(t(n,k).conjugate())
def a(n): return prod(T(n,k) for k in (1..n))
print([a(n).real().round() for n in (0..29)])

Formula

a(n) = Product_{k=1..n} T(n, k) where T(n, k) = t(n,k)*conjugate(t(n,k)) and t(n,k) = 3*cos(Pi*k/n) - i*sin(Pi*k/n), i is the imaginary unit.

a(n) = [x^n] (8*x^3 - 24*x^2 + 6*x + 1)/((4*x - 1)*(2*x + 1)*(x - 1)).

a(n) = n! [x^n] (1 + exp(x) - 2*exp(-2*x) + exp(4*x)).

a(n) = 3*a(n-1) + 6*a(n-2) - 8*a(n-3) for n >= 4.

A062510(n) = sqrt(a(n)) for n > 0.

a(n) = 4^n-2*(-2)^n+1, n>0. - R. J. Mathar, Mar 06 2022