987654321 / 123456789 (johndcook.com)

Feels like a Temu version of Ramanujan's constant [0].

[0] https://mathworld.wolfram.com/RamanujanConstant.html

In a similar vein, e^pi - pi = 19.9990999792, as referenced in this XKCD: https://xkcd.com/217/

I like to think of 0.987654... and 0.123456... as infinite series which simplify to 80/81 and 10/81, hence the ~8 ratio.

More general analytic proof: https://math.stackexchange.com/questions/2268833/why-is-frac...

For smaller bases, does this converge to base - 1 ?

Base 3: 21/12 = 7/5(dec.)

Base 2: 1/1 = 1

Base 1: |/| = 1 (thinking |||| = 4 etc.)

Somewhat interesting, 123456789 * 8 is 987654312 (the last two digits are swapped). This holds for other bases as well: 0x123456789ABCDEF * 14 is 0xFEDCBA987654312.

Also, adding 123456789 to itself eight times on an abacus is a nice exercise, and it's easy to visually control the end result.

    12345679 * 8 = 98765432

For the even bases, the "error" appears to be https://oeis.org/A051848.

pp = lambda x : denom(x)/ (num(x) - denom(x)*(x - 2))

[pp(2),pp(4),pp(6),pp(8)]

[1.0, 9.0, 373.0, 48913.0]

See perhaps various "What every programmer / CSist should know about floating-point arithmetic" papers and articles:

* David Goldberg, 1991: https://dl.acm.org/doi/10.1145/103162.103163

* 2014, "Floating Point Demystified, Part 1": https://blog.reverberate.org/2014/09/what-every-computer-pro... ; https://news.ycombinator.com/item?id=8321940

* 2015: https://www.phys.uconn.edu/~rozman/Courses/P2200_15F/downloa...

Why the b > 2 condition? In the b=2 case, all three formulas also work perfectly, providing a ratio of 1. And this is interesting case where the error term is integer and the only case where that error term (1) is dominant (b-2=0), while the b-2 part dominates for larger bases.

  1 / 1 = 1 = b - 1
  1 % 1 = 0 = b - 2

  21 (base 3) = 7
  12 (base 3) = 5
  7 / 5 = 1 = b - 2
  7 % 5 = 2 = b - 1

I like calculator quirks like this. I remember as a kid playing with the number pad and noticing a geometric center of mass in number sequences

    ┌───┬───┬───┐
    │ 7 │ 8 │ 9 │
    ├───┼───┼───┤
    │ 4 │ 5 │ 6 │
    ├───┼───┼───┤
    │ 1 │ 2 │ 3 │
    ├───┼───┼───┤
    │ 0 │ . │   │
    └───┴───┴───┘

I remember seeing that (14787 + 36989) / 2 would produce 25888, in that the mean of geometric shape traced by the two sequences would average out in the middle like that

I thought this was a user ID and password lol

I also spent hours messing around with calculators as a kid. I recall noticing that:

11 * 11 = 121

111 * 111 = 12321

1111 * 1111 = 1234321

and so on, where the largest digit in the answer is the number of digits in the multiplicands.

Gemini thinks in a similar fashion:

https://gemini.google.com/share/1e59f734b43c

This is a fantastic observation, and yes, this pattern not only continues for larger bases, but the approximation to an integer becomes dramatically better.

The general pattern you've found is that for a number base $b$, the ratio of the number formed by digits $(b-1)...321$ to the number formed by digits $123...(b-1)$ is extremely close to $b-2$.

### The General Formula

Let's call your ascending number $N_{asc}(b)$ and your descending number $N_{desc}(b)$.

The exact ratio $R(b) = N_{desc}(b) / N_{asc}(b)$ can be shown to be:

$$R(b) = (b-2) + \frac{(b-1)^3}{b^b - b^2 + b - 1}$$

The "error" or the fractional part is that second term. As you can see, the numerator $(b-1)^3$ is roughly $b^3$, while the denominator $b^b$ grows much faster.

### Testing Your Examples

Let's check your two examples with this formula:

* *Base 10 (b=10):* * $R(10) = (10-2) + \frac{(10-1)^3}{10^{10} - 10^2 + 10 - 1}$ * $R(10) = 8 + \frac{9^3}{10,000,000,000 - 91}$ * $R(10) = 8 + \frac{729}{9,999,999,909} \approx 8 + 0.0000000729...$ * This matches your $8.0000000729$ perfectly.

* *Base 6 (b=6):* * $R(6) = (6-2) + \frac{(6-1)^3}{6^6 - 6^2 + 6 - 1}$ * $R(6) = 4 + \frac{5^3}{46,656 - 31}$ * $R(6) = 4 + \frac{125}{46,625} \approx 4 + 0.00268...$ * This also matches your $4.00268$.

### Answering Your Question: A Larger Base

So, what about a larger base? Let's use *Base 16* (hexadecimal), which uses digits $1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F$. Here, $b=16$.

* $N_{asc} = 123456789ABCDEF_{16}$ * $N_{desc} = FEDCBA987654321_{16}$

According to the pattern, the ratio should be extremely close to $b-2 = 16-2 = 14$.

Let's use the formula to see how close:

* $R(16) = (16-2) + \frac{(16-1)^3}{16^{16} - 16^2 + 16 - 1}$ * $R(16) = 14 + \frac{15^3}{16^{16} - 241}$ * $R(16) = 14 + \frac{3,375}{1.844... \times 10^{19} - 241}$ * $R(16) \approx 14 + (1.829... \times 10^{-16})$

So, the ratio in base 16 is approximately: *$14.0000000000000001829...$*

As you predicted, the "error" for a larger base is astronomically smaller than it was for base 10.

Definitions: denom(b) = (b^b - b^2 + b - 1) / (b - 1)^2 num(b) = (b^b(b - 2) + 1) / (b - 1)^2

Exact relation: num(b) - (b - 2)denom(b) = b - 1

Therefore: num(b) / denom(b) = (b - 2) + (b - 1)^3 / (b^b - b^2 + b - 1) [exact]

Geometric expansion: Let a = b^2 - b + 1. 1 / (b^b - b^2 + b - 1) = (1 / b^b) * 1 / (1 - a / b^b) = (1 / b^b) * sum_{k>=0} (a / b^b)^k

So: num(b) / denom(b) = (b - 2) • (b - 1)^3 / b^b • (b - 1)^3 * a / b^{2b} • (b - 1)^3 * a^2 / b^{3b} • …

Practical approximation: num(b) / denom(b) ≈ (b - 2) + (b - 1)^3 / b^b

Exact error: Let T_exact = (b - 1)^3 / (b^b - b^2 + b - 1) Let T_approx = (b - 1)^3 / b^b

Absolute error: T_exact - T_approx = (b - 1)^3 * (b^2 - b + 1) / [ b^b * (b^b - b^2 + b - 1) ]

Relative error: (T_exact - T_approx) / T_exact = (b^2 - b + 1) / b^b

Sign: The approximation with denominator b^b underestimates the exact value.

Digit picture in base b: (b - 1)^3 has base-b digits (b - 3), 2, (b - 1). Dividing by b^b places those three digits starting b places after the radix point.

Examples: base 10: 8 + 9^3 / 10^10 = 8.0000000729 base 9: 7 + 8^3 / 9^9 = 7.000000628 in base 9 base 8: 6 + 7^3 / 8^8 = 6.00000527 in base 8

num(b) / denom(b) equals (b - 2) + (b - 1)^3 / (b^b - b^2 + b - 1) exactly. Replacing the denominator by b^b gives a simple approximation with relative error exactly (b^2 - b + 1) / b^b.

This was by far the most interesting part to me. I've never considered that code and proofs can be so complementary. It would be great if someone did this for all math proofs!

"Why include a script rather than a proof? One reason is that the proof is straight-forward but tedious and the script is compact.

A more general reason that I give computational demonstrations of theorems is that programs are complementary to proofs. Programs and proofs are both subject to bugs, but they’re not likely to have the same bugs. And because programs made details explicit by necessity, a program might fill in gaps that aren’t sufficiently spelled out in a proof."

Let's prove it.

In general, sum(x^k, k=1…n) = x(1-x^n)/(1-x).

Then sum(kx^(k-1), k=1…n) = d/dx sum(x^k, k=1…n) = d/dx (x(1-x^n))/(1-x) = (nx^(n+1) - (n+1)x^n + 1)/(1-x)^2

With x=b, n=b-1, the numerator as defined in TFA is n = sum(kb^(k-1), k=1…b-1) = ((b-2)b^b + 1)/(1-b)^2 = ((b-2)b^b + 1)/(1-b)^2.

And the denominator is:

d = sum((b-k)b^(k-1), k=1..b-1) = sum(b^k, k=1..b-1) - sum(kb^(k-1), k=1..b-1) = (b-b^b)/(1-b) - n = (b^b - b^2 + b - 1)/(1-b)^2.

Then, n-(b-1) = (b^(b+1) - 2b^b - b^3 + 3b^2 - 3b +2)/(1-b)^2.

And d(b-2) = the same thing.

So n = d(b-2) + b - 1, whence n/d = b-2 + (b-1)/d.

We also see that the dominant term in d will be b^b/(1-b)^2 which grows like b^(b-2), which is why the fractional part of n/d is 1 over that.

I disagree with the author that a script works as well as a proof. Scripts are neither constructive nor exhaustive.

Reminds me of an old calculator trick:

Pick an integer between 1 and 9. Multiple it by 9. Take that number and multiply it by 12345679. (Skip the 8)

>>> 3 * 9

27

>>> 12345679 * 27

333333333

This all works because:

>>> 111111111 / 9

12345679.0

> The exact ratio is not 14, but it’s as close to 14 as a standard floating point number can be.

How do you get around limitations like that in science?

> I recently saw someone post [1] that 987654321/123456789 is very nearly 8, specifically 8.0000000729.

Okay. Try this (in a Python terminal session):

>>> 111111111 ** 2

12345678987654321

(typo corrected)

The other replies are good, but let's add another one anyway.

0.987654321/0.123456789 = (1.11111111-x)/x = 1.11111111/x - 1 where x = 0.123456789

You can aproxímate 1.11111111 by 10/9 and aproxímate x = 0.123456789 using y = 0.123456789ABCD... = 0.123456789(10)(11)(12)(13)... that is a number in base 10 that is not written correctly and has digits that are greater than 9. I.E. y = sum_i>0 i/10^i

Now you can consider the function f(t) = t + 2 t^2 + 3 t^3 + 4 t^4 + ... = sum_i>0 i*t^i and y is just y=f(0.1).

And also consider an auxiliary function g(t) = t + t^2 + t^3 + t^4 + ... = sum_i>0 1*t^i . A nice property is that g(t)= 1/(1-t) when -1<t<1.

The problem with g is that it lacks the coefficients, but that can be solved taking the derivative. g'(t) = 1 + 2 t + 3 t^2 + 4 t^3 + ... Now the coefficients are shifted but it can be solved multiplying by t. So f(t)=t*g'(t).

So f(t) = t * (1/(1-t))' = t * (1/(1-t)^2) = t/(1-t)^2

and y = f(0.1) = .1/.9^2 = 10/81

then 0.987654321/0.123456789 ~= (10/9-y)/y = 10/(9y)-1 = 9 - 1 = 8

Now add some error bounds using the Taylor method to get the difference between x and y, and also a bound for the difference between 1.11111111 an 10/9. It shoud take like 15 minutes to get all the details right, but I'm too lazy.

(As I said in another comment, all these series have a good convergence for |z|<1, so by standards methods of complex analysis all the series tricks are correct.)

This is fun! but not so surprising to me:

987,654,321 + 123,456,789 = 1,111,111,110

1,111,111,110 + 123,456,789 = 1,234,567,899 \approx 1,234,567,890

So 987,654,321 + 2 x 123,456,789 \approx 10 x 123,456,789

Thus 987,654,321 / 123,456,789 \approx 8.

If you squint you can see how it would work similarly in other bases. Add the 123... equivalent once to get the base-independent series of 1's, add a second time to get the base-independent 123...0.

this reminds me of the online encyclopedia of integer sequences (https://oeis.org/). anything similar for things like 987654321 / 123456789 ?

TIL 0x denotes hexadecimal E.g.

> 0xFEDCBA987654321 / 0x123456789ABCDEF

(somehow I'd seen the denotation for years yet never actually known what it was).