Necklace conjectures involving $f(n) =$ the number of $n-$bead necklaces with $2$ colors. (A000031)

Question

consider $f(n)$ for $n>0$ :

$f(n) =$ the number of $n-$bead necklaces with $2$ colors when turning over is not allowed.

So we get

$$f(1) = 2$$

(a binary bit or bead)

$$f(2) = 3$$

($11$,$00$ or $01 = 10$ so $3$ ways)

$$f(3) = 4$$

$$f(4) = 6$$

We get the list

$$2, 3, 4, 6, 8, 14, 20, 36, 60, 108, 188, 352, 632, 1182, 2192, ...$$

basically equal to

https://oeis.org/A000031

Conjecture $C$ :

for all $n>0$ :

$$f(n) < 2^{L n} + n$$

Where $L$ satisfies $L + \exp(L) = 2$

So $L = 2 - W(e^2)$

($L$ is about $0.4428$)

where $W(x)$ is the LambertW function. ( https://en.wikipedia.org/wiki/Lambert_W_function )

Conjecture $D$ :

$$ \lim \frac{\ln(f(n))}{\ln(2) n} = L^*$$

for some constant $L^*$.

Can we prove those ?

What is the value of $L^*$ ??

Do we have $L^* = L $ ?

It seems $L^*$ is at most $L$.

Another candidate for $L^*$ is $V$

$$ L^* =^? V = \int_0^{\infty} \frac{dx}{1 + 2^x + 3^x} $$

Where $V$ seems like a lower bound for $L^*$.

($V$ is about $0.4316$ )

Im not sure how Pólya enumeration theorem and the alike can help here if at all. Same for the Necklace polynomials.

I am not sure what are the open problems or solved ones regarding this.

That weird integral occured in the comments here today :

Integral $\int_1^\infty\frac{dx}{1+2^x+3^x}$

where user https://math.stackexchange.com/users/905886/%d0%a2yma-gaidash

considered changing the domain of integration.

I recognized that value from the numerical experiments I did recently with A000031, although none of that is formal or anything.

I was unable to find anything in the books that seemed like helpful to me, although I am new to this.

Answer 1

Conjecture C is false. It's a standard exercise using Burnside's lemma to show that

$$f(n) = \frac{1}{n} \sum_{d \mid n} \varphi(d) 2^{\frac{n}{d}}$$

which gives

$$f(n) \ge \frac{2^n}{n}.$$

So no inequality of the form $f(n) < 2^{Ln} + n$ can hold for any $L < 1$. I have no idea where your value of $L$ comes from numerically; with $L = 0.4428$ your conjecture already fails for $n = 6$.

Conjecture D is true with $L^{\ast} = 1$, since we also have an asymptotically matching upper bound $f(n) \le 2^n$, which gives

$$\lim_{n \to \infty} \frac{\ln f(n)}{n \ln 2} = 1.$$

I have no idea what $V$ has to do with this sequence.