A032087 Number of reversible strings with n beads of 4 colors. If more than 1 bead, not palindromic.

4, 6, 24, 120, 480, 2016, 8064, 32640, 130560, 523776, 2095104, 8386560, 33546240, 134209536, 536838144, 2147450880, 8589803520, 34359607296, 137438429184, 549755289600, 2199021158400, 8796090925056, 35184363700224, 140737479966720, 562949919866880

Offset: 1

Author

Christian G. Bower

Keywords

• nonn
• easy

Comments

From Petros Hadjicostas, Jun 30 2018: (Start)

Using the formulae in C. G. Bower's web link below about transforms, it can be proved that, for k >= 2, the BHK[k] transform of sequence (c(n): n >= 1), which has g.f. C(x) = Sum_{n >= 1} c(n)*x^n, has generating function B_k(x) = (1/2)*(C(x)^k - C(x^2)^{k/2}) if k is even, and B_k(x) = C(x)*B_{k-1}(x) = (C(x)/2)*(C(x)^{k-1} - C(x^2)^{(k-1)/2}) if k is odd. For k=1, Bower assumes that the BHK[k=1] transform of (c(n): n >= 1) is itself, which means that the g.f. of the output sequence is C(x). (This assumption is not accepted by all mathematicians because a sequence of length 1 is not only reversible but palindromic as well.)

Since a(m) = BHK(c(n): n >= 1)(m) = Sum_{k=1..m} BHK[k](c(n): n >= 1)(m) for m = 1,2,3,..., it can be easily proved (using sums of infinite geometric series) that the g.f. of BHK(c(n): n >= 1) is A(x) = (C(x)^2 - C(x^2))/(2*(1-C(x))*(1-C(x^2))) + C(x). (The extra C(x) is due of course to the special assumption made for the BHK[k=1] transform.)

Here, BHK(c(n): n >= 1)(m) indicates the m-th element of the output sequence when the transform is BHK and the input sequence is (c(n): n >= 1). Similarly, BHK[k](c(n): n >= 1)(m) indicates the m-th element of the output sequence when the transform is BHK[k] (i.e., with k boxes) and the input sequence is (c(n): n >= 1).

For the current sequence, c(1) = 4, and c(n) = 0 for all n >= 2, and thus, C(x) = 4*x. Substituting into the above formula for A(x), and doing the algebra, we get A(x) = 2*x*(2-5*x-8*x^2+32*x^3) / ((1+2*x)*(1-2*x)*(1-4*x)), which is R. J. Mathar's formula below.

(End)

The formula for a(n) for this sequence was Ralf Stephan's conjecture 72. It was solved by Elizabeth Wilmer (see Proposition 1 in one of the links below). She does not accept Bower's assertion that a string of length 1 is not palindromic. - Petros Hadjicostas, Jul 05 2018

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• C. G. Bower, Transforms (2)
• Ralf Stephan, Prove or disprove: 100 conjectures from the OEIS, arXiv:math/0409509 [math.CO], 2004.
• Elizabeth Wilmer, Notes on Stephan's conjectures 72, 73 and 74
• Elizabeth Wilmer, Notes on Stephan's conjectures 72, 73 and 74 [cached copy]
• Index entries for linear recurrences with constant coefficients, signature (4,4,-16).

Crossrefs

Column 4 of A293500 for n>1.

Cf. A000302, A026337 (bisection), A032121, A056450, A088037.

Programs

Magma

A032087:= func< n | n eq 1 select 4 else 2^(2*n-1) -(3-(-1)^n)*2^(n-2) >;
[A032087(n): n in [1..30]]; // G. C. Greubel, Oct 02 2024

Mathematica

Join[{4}, LinearRecurrence[{4, 4, -16}, {6, 24, 120}, 24]] (* Jean-François Alcover, Oct 11 2017 *)

PARI

Vec(2*x*(2 - 5*x - 8*x^2 + 32*x^3) / ((1 - 2*x)*(1 + 2*x)*(1 - 4*x)) + O(x^30)) \\ Colin Barker, Mar 08 2017

SageMath

def A032087(n): return 2^(2*n-1) -3*2^(n-2) +(-2)^(n-2) +4*int(n==1)
[A032087(n) for n in range(1,31)] # G. C. Greubel, Oct 02 2024

Formula

"BHK" (reversible, identity, unlabeled) transform of 4, 0, 0, 0, ...

a(2*n+1) = 2^(4*n+1) - 2^(2*n+1), a(2*n) = 2^(4*n-1) - 2^(2*n) + 2^(2*n-1), a(1)=4.

a(n) = (A000302(n) - A056450(n))/2 for n > 1.

From R. J. Mathar, Mar 20 2009: (Start)

a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) for n > 4.

G.f.: 2*x*(2-5*x-8*x^2+32*x^3)/((1-2*x)*(1+2*x)*(1-4*x)). (End)

From Colin Barker, Mar 08 2017: (Start)

a(n) = 2^(n-1) * (2^n-1) for n > 1 and even.

a(n) = 2^(2*n-1) - 2^n for n > 1 and odd. (End)

E.g.f.: (1/4)*( exp(-2*x) - 3*exp(2*x) + 2*exp(4*x) ) + 4*x. - G. C. Greubel, Oct 02 2024