The Fibonacci sequence $P_n = P_{n-1}+P_{n-2}$ is
$$1,1,2,3,5,8,13,21,34,55,89,144,233,377, 610, \cdots $$
I learnt that the fraction $1/89$ contains all the numbers in the sequence.
$$\begin{align}
\frac{1}{89}&= 0.\overline{01123595505617977528089887640449438202247191~}\\
&=0.01+0.001+0.0002+0.00003+0.000005+0.0000008+~\\
&~~~~~0.00000013+0.000000021+0.0000000034+0.00000000055+ ~\\
&~~~~~0.000000000089+0.0000000000144+0.00000000000233+~\\
&~~~~~0.000000000000377+0.0000000000000610+\cdots
\end{align}$$
where the over line represents repeated cycle.
Rule of the number of zeros (not sure if this is right or not):
Don't add zero for the next number if it is "smaller" than the previous number. To compare numbers, we only keep the first digit and make the rest of the digits after the decimals point. For example, $13$ in this case is "smaller" than $8$ because $1.3<8$, so we don't add any zero for $13$ — the same $7$ zeros in front of both $13$ and $8$. On the other hand, if the number in the sequence is greater than or equal to the previous one, we would add a zero before the greater number. For example, $3>2$, so we add a zero in front of $3$, making $5$ zeros in front of $3$ and $4$ zeros in front of $2$.
I think that the rule of the number of zeros applies to all the metallic sequences. If not, let's assume it is for now and keep on reading.
I then decided to further explore other metallic sequences. Lets define the $n^{th}$ metallic sequence
$$\sigma_n: P_n = nP_{n-1}+P_{n-2}$$
In this post, the Fibonacci sequence is $\sigma_1$.
The next metallic sequence $\sigma_2$, or the silver sequence, is
$$\sigma_2: P_n = 2P_{n-1}+P_{n-2}$$
$$1,2,5,12,29,70,169,408,985,2378,5741,13860,33461,80782,\cdots$$
I took a guess that $1/79$ would contain all the numbers in $\sigma_2$, and it seems like I am correct for the numerical value although I am not sure how to prove the relationship.
$$\begin{align}
\frac{1}{79}&=0.\overline{0126582278481}\\
&= 0.01+0.002+0.005+0.00012+0.000029+0.0000070+~\\
&~~~~~0.00000169+0.000000408+0.0000000985+~\\
&~~~~~0.00000002378+0.000000005741+0.000000001386+~\\
&~~~~~0.00000000033461+0.000000000080782+\cdots
\end{align}$$
I will present two more cases so that you will get the idea of the pattern.
Here is $\sigma_3$, or copper sequence:
$$\sigma_3: P_n = 3P_{n-1}+P_{n-2}$$
$$1,3,10,33,109,360,1189,3927,12970,42837,141481,467280$$
$$\begin{align}
\frac{1}{69}&=0.\overline{01449275362}\\
&= 0.01+0.003+0.0010+0.00033+0.000109+0.0000360+~\\
&~~~~~0.00001189+0.000003927+0.0000012970+~\\
&~~~~~0.00000042837+0.000000141481+0.000000046728+~\cdots
\end{align}$$
Lastly, I will present the case for $\sigma_{9}$:
$$\sigma_9: P_n = 9P_{n-1}+P_{n-2}$$
$$1,9,82,747,6805,61992,564733,5144589,46866034,426938895,3889316089,\cdots$$
$$\begin{align}
\frac{1}{9}&=0.\overline{1}\\
&=0.01+0.009+0.0082+0.00747+0.006805+0.0061992+~\\
&~~~~~0.00564733+0.005144589+0.0046866034+0.00426938895+~\\
&~~~~~0.003889316089+\cdots
\end{align}$$
For $\sigma_9$, I know that if you only type these numbers into the calculator, the value is by no way close to $1/9$ because the series approaches $1/9$ very slowly, so we have to type in a lot of numbers to get the value close to $1/9$.
Now, I have two questions on hand:
$1)$ How to prove that a fraction, such as $1/89,~1/79,~1/69,\cdots,~1/9$, is the sum of all the numbers in the corresponding metallic sequence?
$2)$ I am trying to find a fraction that contains all the numbers in $\sigma_{10}$, but with no avail. Are there any other fractions that contain all the numbers in the metallic sequence $\sigma_{10}$? Maybe also fractions for $\sigma_{11},~ \sigma_{12}$, and so on?
Best Answer
Answer to question 1) :
The generating function for the Fibonacci numbers $F_n$ is known to be
$$\dfrac{1}{1-(x+x^2)}=\underbrace{1}_{F_0}+\underbrace{1}_{F_1}x+\underbrace{2}_{F_2}x^2+\underbrace{3}_{F_3}x^3+\underbrace{5}_{F_4}x^4+\cdots+F_nx^n+...$$
Taking $x=0.1$ gives :
$$\dfrac{1}{1-0.11}=1+1 \times 0.1+2 \times 0.01+3 \times 0.001+5 \times 0.0001+\cdots+F_n 0.1^n+...$$
justifying the equality of LHS and RHS of your first identity (multiplied by $100$).
Same process for the other metallic sequences.
For example, the generating functions of the silver and bronze sequences are resp.
$$\dfrac{1}{1-(2x+x^2)} \ \ \ \text{and} \ \ \ \dfrac{1}{1-(3x+x^2)}$$
An interesting generalization along these lines : the recent paper https://arxiv.org/pdf/1901.02619.pdf
Remark: one finds as well this generating function for the Fibonacci numbers:
$$\dfrac{x}{1-x-x^2}$$
if one takes a shifted-by-one convention:
$$F_0=0, \ \ F_1=1, \ \ , F_2=1, \ \ F_3=2, \cdots$$
(this is the convention taken in OEIS)