Of course I cannot prove this, but I am pretty sure that God's number for the nxnxn cube is

C(n)*n^2/Log(n) with

C(n)->0.25Log(24!)-1.5Log(4!) for n->Infinity

Maybe it's a coincidence, but when I looked at my

derivative formula a second time (

from this post), I noticed something interesting.

You're saying that \( C\left( n \right)=\frac{1}{4}\ln \left( 24! \right)-\frac{3}{2}\ln \left( 4! \right)=\ln \left( \frac{24!^{1/4}}{4!^{3/2}} \right) \).

The only two non constant factors in the derivative formula besides the natural log factor happen to be: \( \frac{24!^{\left( 1/4 \right)\left( n+1 \right)\left( n-3 \right)}}{4!^{\left( 3/2 \right)\left( n-1 \right)\left( n-3 \right)}} \).

So \( \ln \left( \frac{24!^{\frac{1}{4}\left( n+1 \right)\left( n-3 \right)}}{4!^{\frac{3}{2}\left( n-1 \right)\left( n-3 \right)}} \right)\frac{n^{2}}{\ln \left( n \right)} \)

The command "Log[24!^((1/4) (Range[100] + 1) (Range[100] - 3))/4!^((3/2) (Range[100] - 1) (Range[100] - 3))] (Range[100]^2/Log[Range[100]])//N" in Mathematica yields (values for n=1 to n=100):

{ComplexInfinity , -209.603 , 0. , 625.318 , 1960.58 , 4342.15 , 8155.32 , 13833.9 , 21859. , 32758. , 47103.6 , 65513. , 88647.1 , 117210. , 151947. , 193647. , 243139. , 301293. , 369021. , 447272. , 537038. , 639348. , 755271. , 885915. , 1.03243*10^6 , 1.19599*10^6 , 1.37782*10^6 , 1.57919*10^6 , 1.80138*10^6 , 2.04574*10^6 , 2.31364*10^6 , 2.60648*10^6 , 2.9257*10^6 , 3.2728*10^6 , 3.64929*10^6 , 4.05671*10^6 , 4.49667*10^6 , 4.97078*10^6 , 5.48071*10^6 , 6.02816*10^6 , 6.61484*10^6 , 7.24254*10^6 , 7.91306*10^6 , 8.62822*10^6 , 9.38991*10^6 , 1.02*10^7 , 1.10605*10^7 , 1.19733*10^7 , 1.29405*10^7 , 1.39641*10^7 , 1.50462*10^7 , 1.61888*10^7 , 1.73942*10^7 , 1.86645*10^7 , 2.00019*10^7 , 2.14087*10^7 , 2.28872*10^7 , 2.44396*10^7 , 2.60684*10^7 , 2.77758*10^7 , 2.95644*10^7 , 3.14365*10^7 , 3.33947*10^7 , 3.54414*10^7 , 3.75792*10^7 , 3.98107*10^7 , 4.21384*10^7 , 4.45652*10^7 , 4.70935*10^7 , 4.97261*10^7 , 5.24658*10^7 , 5.53154*10^7 , 5.82775*10^7 , 6.13552*10^7 , 6.45512*10^7 , 6.78685*10^7 , 7.13099*10^7 , 7.48785*10^7 , 7.85773*10^7 , 8.24093*10^7 , 8.63775*10^7 , 9.0485*10^7 , 9.4735*10^7 , 9.91307*10^7 , 1.03675*10^8 , 1.08372*10^8 , 1.13223*10^8 , 1.18234*10^8 , 1.23406*10^8 , 1.28744*10^8 , 1.3425*10^8 , 1.39928*10^8 , 1.45781*10^8 , 1.51814*10^8 , 1.58029*10^8 , 1.6443*10^8 , 1.7102*10^8 , 1.77804*10^8 , 1.84784*10^8 , 1.91965*10^8}

(For those who may be wondering, "complex infinity" in Mathematica just means dividing by zero).

If we start with the value for n=4, that is, 625.318 and divide it by the value for n=3 (well, in this case it would be 20 and not zero), we get 31.2659. Likewise, if we continue interating with 1960.58/31.2659 =62.70665485 for the 5x5x5, and so on, we get:

[FONT="]

Code:

```
n=2: 11
n=3: 20
n=4: 31.2659
n=5: 62.70665485
n=6: 69.24544149
n=7: 117.7741065
n=8: 117.4613029
n=9: 186.0953306
n=10: 176.0280599
n=11: 267.5914284
n=12: 244.8247329
n=13: 362.0839241
n=14: 323.7094834
n=15: 469.3931065
n=16: 412.547601
n=17: 589.3598688
n=18: 511.2207599
n=19: 721.8427516
n=20: 619.6252563
n=21: 866.7141865
n=22: 737.6687839
n=23: 1 023.861951
n=24: 865.2680173
n=25: 1 193.190987
n=26: 1 002.345821
n=27: 1 374.595445
n=28: 1 148.839832
n=29: 1 567.999255
n=30: 1 304.681743
n=31: 1 773.336688
n=32: 1 469.816769
n=33: 1 990.520222
n=34: 1 644.193294
n=35: 2 219.501815
n=36: 1 827.75701
n=37: 2 460.212149
n=38: 2 020.468033
n=39: 2 712.594266
n=40: 2 222.285904
n=41: 2 976.592701
n=42: 2 433.164604
n=43: 3 252.167974
n=44: 2 653.067145
n=45: 3 539.265871
n=46: 2 881.953595
n=47: 3 837.848056
n=48: 3 119.795215
n=49: 4 147.868404
n=50: 3 366.572571
n=51: 4 469.29323
n=52: 3 622.228207
n=53: 4 802.071821
n=54: 3 886.75986
n=55: 5 146.163056
n=56: 4 160.128579
n=57: 5 501.560725
n=58: 4 442.303052
n=59: 5 868.217385
n=60: 4 733.260235
n=61: 6 246.09646
n=62: 5 032.839342
n=63: 6 635.359829
n=64: 5 341.292848
n=65: 7 035.600007
n=66: 5 658.465512
n=67: 7 446.96595
n=68: 5 984.343194
n=69: 7 869.451746
n=70: 6 318.8773
n=71: 8 303.025603
n=72: 6 662.077494
n=73: 8 747.646669
n=74: 7 013.909263
n=75: 9 203.312673
n=76: 7 374.355562
n=77: 9 669.983959
n=78: 7 743.394437
n=79: 10 147.65561
n=80: 8 121.018605
n=81: 10 636.28889
n=82: 8 507.196532
n=83: 11 135.86593
n=84: 8 901.930092
n=85: 11 646.35073
n=86: 9 305.232389
n=87: 12 167.67032
n=88: 9 717.06143
n=89: 12 699.93
n=90: 10 137.37871
n=91: 13 243.06844
n=92: 10 566.13131
n=93: 13 797.00818
n=94: 11 003.40002
n=95: 14 361.83359
n=96: 11 449.0952
n=97: 14 937.42493
n=98: 11 903.25648
n=99: 15 523.81908
n=100: 12 365.83595
```

[/FONT]By these "results," God's number would appear to be higher in odd cubes. Specifically, God's number for n=7 and n=8 would be nearly identical, but then the values for odd cubes surpasses that of the even cubes henceforth.

Do these results seem to be reasonable? I can tell that for n=100, the result here is less than the approximation of the lowerbound you posted by about 1000.