**A Simple Algorithm for Approximating God's Number**
Hey guys,

The following is just another one of my crazy approximations of god's number (I think it works best with

h-b move metric, which I think is OBTM), but it is based off of such a simple idea, and the results seem to be very reasonable, that I thought I should just post it before I forget it.

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**Algorithm**

Let's let a formula which gives the number of positions of the

*n*x

*n*x

*n* cube be called \( f\left( n \right) \), then I "claim" that god's number is within a small proximity of the result from using the following algorithm.

[1] Choose a cube size for which you want to approximate god's number for in OBTM and call it N.

[2] Evaluate \( \left\lfloor \text{Log}_{10}\left[ f\left( N \right) \right]+0.5 \right\rfloor \) to get the number of digits of \( f\left( N \right) \).

[3] Now, by experimentation, substitute values for

*k* in \( \left\lfloor \text{Log}_{10}\left[ k! \right]+0.5 \right\rfloor \) to try to match the value calculated in step [2]. Choose

*k* + 1 if

*k* is smaller than the result from [2] and

*k* + 1 is larger than the result from [2].

The God's number approximation will be the value of

*k* which is obtained in step [3].

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**Formula**

Using

this approximation of the number of digits of

*k*! and using Mathematica, I get the following formula which "calculates the lower and upperbound" for a cube of size N.

\( \text{God }\!\!'\!\!\text{ sNo.}\left( n \right)=\left\lfloor \frac{\left\lfloor \text{Log}_{10}\left[ f\left( n \right) \right]+0.5 \right\rfloor \left( \ln \left( 2 \right)+\ln \left( 5 \right) \right)}{\text{ProductLog}\left[ \frac{1}{b}\left\lfloor \text{Log}_{10}\left[ f\left( n \right) \right]+0.5 \right\rfloor \left( \ln \left( 2 \right)+\ln \left( 5 \right) \right) \right]} \right\rfloor \)

, where:

ProductLog[x] is the inverse function of \( xe^{x} \),

choose \( b=2 \) to get the "lowerbound" and choose \( b=3 \) to get the "upperbound".

, and again, \( f\left( n \right) \) is any number of positions formula for cube size

*n*.

This formula is only necessary for very large

*n*, and its purpose is to give you an interval where to start testing

*k* values when finding the number of digits

*k*! has (which is step [3] of the algorithm).

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**Results from the Algorithm and the Formula**

Below, "GN" is the "actual" God's number approximation using the algorithm, and the bracketed pair is the lowerbound and upperbound achieved from the formula.

n=2, {10,11}, GN = 10

n=3, {20,22}, GN = 21

n=4, {36,40}, GN = 39

n=5, {52,57}, GN = 56

n=6, {73,81}, GN = 79

n=7, {95,103}, GN = 101

n=8, {122,132}, GN = 129

n=9, {148,160}, GN = 157

n=10, {179,193}, GN = 189

n=11, {210,226}, GN = 222

n=12, {245,263}, GN = 259

n=13, {281,301}, GN = 296

n=14, {321,343}, GN = 338

n=15, {360,385}, GN = 379

n=16, {404,432}, GN = 425

n=17, {448,478}, GN = 471

n=18, {496,529}, GN = 521

n=19, {544,580}, GN = 571

n=20, {596,635}, GN = 625

n=100, {10344,10801}, GN = 10687

n=1000, {698895,720084}, GN = 714818

n=1001, {700192,721417}, GN = 716142

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**Comment**

I can see that my approximation for

*n* = 1001 is smaller than

qqwref's lowerbound of 981766, but I'm not sure exactly if his was correct.