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edited Sep 28 at 0:20

To make it less surprising is to fade some of the magic! But, ok.

First let us say that any real sequence $a_1,a_2,\cdots$ has an ordinary generating function (ogf) $f(x)=\sum a_ix^i$ which may be a formal series with radius of convergence $0$ (and still be useful) BUT if the $a_i$ are positive integers and $f(x)$ converges at $\frac{1}{b^k}$ then $f(\frac{1}{b^k})$ is a number whose base $b$ expansion is the sequence $a_i$ buffered by $0$'s until they start to bump into each other. I'll stick to $b=10$ and I'll use $(10^{-j})f(10^{-k})$ if there is an $a_0$ term I want to shift past the decimal point.

So the question might be which series have a nice ogf? That the Catalan numbers do is very nice.

From $\frac{1}{(1-x)^k}=\sum \binom{k+i}{k}x^i$ one obtains

$\frac{1}{0.9998}=0.10002000400080016003200640128025605121024$

$\frac{1}{10(0.999)^2}=0.10002000300040005000600070008000900100011$

$\frac{1}{10(0.999)^3}=0.10003000600100015002100280036004500550066$

Since $\binom{i}{2}+\binom{i+1}{2}=i^2$,

we can use $(\frac{1}{10}+\frac{1}{100000})\frac{1}{(1-x)^3}$ at $x=0.0001$ to get

$\frac{1.0001}{10(0.999)^3}=\frac{100010000000}{999700029999}=0.100040009001600250036004900640081010001210$

But it is more productive to use $\binom{i}{1}+2\binom{i}{2}=i^2$ for

$\frac{1}{10(1-x)^2}+\frac{1}{5000(1-x)^3}.$ Another function which coincides with the previous one at $x=0.0001.$ So the same rational but obtained another way. This second approach makes it clear how to get any polynomial sequence $a_i=p(i).$

This already seems less fun. So thinking of a dramatic last target, the Fibonacci numbers remind us that anything us that anything given by a recurrence relation (linear, with constant coefficients...) has a nice ogf.

At $x=\frac{1}{100}$,

$\frac{1}{10(1-x-x^2)}=\frac{100000}{998999}=0.1001002003005008013021034055089144$

A cute point is that all my examples (none of which are as nice as yours) lead to a rational number so as the $a_i$ increase in size and overlap they result in an eventually repeating decimal. For example the base $10$ "Fibonacci" rational I gave has period $496620$ while $\frac{10}{89}=0.\overline{11235955056179775280898876404494382022471910}$

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edited Sep 27 at 17:33

Aaron Meyerowitz
11.6k●1●10●31

To make it less surprising is to fade some of the magic! But, ok.

First let us say that any real sequence $a_1,a_2,\cdots$ has an ordinary generating function (ogf) $f(x)=\sum a_ix^i$ . $f(x)$ which may be a formal series with radius of convergence $0$ (and still be useful) BUT if the $a_i$ are positive integers and $f(x)$ converges at $\frac{1}{b^k}$ then $f(\frac{1}{b^k})$ is a number whose base $b$ expansion is the sequence $a_i$ buffered by $0$'s until they start to bump into each other. I'll stick to $b=10$ and I'll use $(10^{-j})f(10^{-k})$ if there is an $a_0$ term I want to shift past the decimal point.

So the question might be which series have a nice ogf? That the Catalan numbers do is very nice.

From $\frac{1}{(1-x)^k}=\sum \binom{k+i}{k}x^i$ one obtains

$\frac{1}{0.9998}=0.10002000400080016003200640128025605121024$

$\frac{1}{10(0.999)^2}=0.10002000300040005000600070008000900100011$

$\frac{1}{10(0.999)^3}=0.10003000600100015002100280036004500550066$

Since $\binom{i}{2}+\binom{i+1}{2}=i^2$,

we can use $(\frac{1}{10}+\frac{1}{100000})\frac{1}{(1-x)^3}$ at $x=0.0001$ to get

$\frac{1.0001}{10(0.999)^3}=0.100040009001600250036004900640081010001210$\frac{1.0001}{10(0.999)^3}=\frac{100010000000}{999700029999}=0.100040009001600250036004900640081010001210$

But it is more productive to use $\binom{i}{1}+2\binom{i}{2}=i^2$ for

$(\frac{1}{10}\frac{1}{(1-x)^2}+\frac{1}{50000})\frac{1}{(1-x)^3}$ \frac{1}{10(1-x)^2}+\frac{1}{5000(1-x)^3}.$ Another function which coincides with the previous one at $x=0.0001$ so x=0.0001.$ So the same rational but obtained another way. This second approach makes it clear how to get any polynomial sequence $a_i=p(i).$

At $x=\frac{1}{100}$,

$\frac{1}[10(1-x-x^2)}=\frac{100000}{998999}=0.1001002003005008013021034055089144$\frac{1}{10(1-x-x^2)}=\frac{100000}{998999}=0.1001002003005008013021034055089144$

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answered Sep 27 at 17:27

Aaron Meyerowitz
11.6k●1●10●31

To make it less surprising is to fade some of the magic! But, ok.

First let us say that any real sequence $a_1,a_2,\cdots$ has an ordinary generating function (ogf) $f(x)=\sum a_ix^i$. $f(x)$ which may be a formal series with radius of convergence $0$ (and still be useful) BUT if the $a_i$ are positive integers and $f(x)$ converges at $\frac{1}{b^k}$ then $f(\frac{1}{b^k})$ is a number whose base $b$ expansion is the sequence $a_i$ buffered by $0$'s until they start to bump into each other. I'll stick to $b=10$ and I'll use $(10^{-j})f(10^{-k})$ if there is an $a_0$ term I want to shift past the decimal point.

So the question might be which series have a nice ogf? That the Catalan numbers do is very nice.

From $\frac{1}{(1-x)^k}=\sum \binom{k+i}{k}x^i$ one obtains

$\frac{1}{0.9998}=0.10002000400080016003200640128025605121024$

$\frac{1}{10(0.999)^2}=0.10002000300040005000600070008000900100011$

$\frac{1}{10(0.999)^3}=0.10003000600100015002100280036004500550066$

Since $\binom{i}{2}+\binom{i+1}{2}=i^2$,

we can use $(\frac{1}{10}+\frac{1}{100000})\frac{1}{(1-x)^3}$ at $x=0.0001$ to get

$\frac{1.0001}{10(0.999)^3}=0.100040009001600250036004900640081010001210$

But it is more productive to use $\binom{i}{1}+2\binom{i}{2}=i^2$ for

$(\frac{1}{10}\frac{1}{(1-x)^2}+\frac{1}{50000})\frac{1}{(1-x)^3}$ Another function which coincides with the previous one at $x=0.0001$ so the same rational but obtained another way. This second approach makes it clear how to get any polynomial sequence $a_i=p(i).$

At $x=\frac{1}{100}$,

$\frac{1}[10(1-x-x^2)}=\frac{100000}{998999}=0.1001002003005008013021034055089144$