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.

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 │ . │   │
    └───┴───┴───┘

This was by far the most interesting part to me. I've never considered that code and proofs can be so complimentary. 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."

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.

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.

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.

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...

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]

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

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

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

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.

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.)

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

I thought this was a user ID and password lol

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.