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?

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    $\begingroup$ This is a related post, but the generalization is towards the tribonacci, tetranacci, etc $\endgroup$ Commented Jun 23, 2019 at 14:28
  • $\begingroup$ Ok, I see, thanks! $\endgroup$
    – Larry
    Commented Jun 23, 2019 at 20:43

2 Answers 2


Answer to question 1) :

The generating function for the Fibonacci numbers $F_n$ is known to be


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:


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)

  • $\begingroup$ Thanks for the link to this interesting paper. $\endgroup$ Commented Sep 10, 2019 at 6:55

Following on from Jean Marie's answer, the metallic sequence $$M_{n,k}=nM_{n,k-1}+M_{n,k-2}$$ Has the generating function $$G_n(x)=M_{n,0}+M_{n,1}x+M_{n,2}x^2+\dots$$ Such that $$xG_n(x)=M_{n,0}x+M_{n,1}x^2+M_{n,2}x^3+\dots$$ $$nG_n(x)=nM_{n,0}+nM_{n,1}x+nM_{n,2}x^2+\dots$$ $$(x+n)G_n(x)=nM_{n,0}+(nM_{n,1}+M_{n,0})x+(nM_{n,2}+M_{n,1})x^2+\dots$$ $$(x+n)G_n(x)=nM_{n,0}+M_{n,2}x+M_{n,3}x^2+\dots$$ $$(x+n)G_n(x)=nM_{n,0}+\frac{G_n(x)-M_{n,0}}x-M_{n,1}$$ $$x(x+n)G_n(x)=nM_{n,0}x+G_n(x)-M_{n,0}-M_{n,1}x$$ $$(x(x+n)-1)G_n(x)=(nM_{n,0}-M_{n,1})x-M_{n,0}$$ $$G_n(x)=\frac{M_{n,0}+(M_{n,1}-nM_{n,0})x}{1-x(x+n)}$$ But we have the values of $M_{n,0}=0$ and $M_{n,1}=1$ hence this just becomes $$G_n(x)=\frac{x}{1-x(x+n)}$$ If we let $x=\frac1{10}$ we get the fractional representation as mentioned, $$G_n\left(\frac1{10}\right)=\frac{1/10}{1-(1/10+n)/10}=\frac{10}{99-10n}$$ Which gives fractional values of $$\frac{10}{89},\frac{10}{79},\frac{10}{69},\frac{10}{59},\dots$$ and each of these still contain the corresponding metallic sequence as, for example, $$\frac{10}{89}=0.\overline{11235955056179775280898876404494382022471910}$$

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    $\begingroup$ [+1] Thanks for having given a detailed constructive proof of the general generating function of all metallic sequences (I just gave 2 examples, without proof). $\endgroup$
    – Jean Marie
    Commented Jun 14, 2019 at 22:58
  • $\begingroup$ It is interesting to note that the given period of the decimal expansion of $10/89$ has length $44$, a divider of $\varphi(p)=p-1=88$ (Euler totient function) with $p=89$; see answers to this question. $\endgroup$
    – Jean Marie
    Commented Jun 30, 2019 at 7:26

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