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A day or two ago, PsiCubed2 created an ever-growing googolism, or PEGG for short. I think this is a great idea, and I wanted to try out the idea a bit further, and wanted to know what will happen to PEGG in the future. So I decided to simulate it.

Since I don't think I can predict future values of Dow Jones, I slightly modified the definition of PEGG for our simulation purposes. Specifically, rule (a) in the definition is replaced with the following:

(a) We take the (days passed since 2017-05-08)-th integer in the result from the Random Integer Generator by random.org, where the minimum value is 1, the maximum value is 10, and the pregenerated randomization from 2017-05-08 is used (you'll need to switch to Advanced Mode to set this). Call it n.

The first 100 integers produced using the generator are:

10	2	8	8	3
2	9	4	9	5
7	3	8	4	4
8	5	9	3	9
8	2	8	7	2
5	5	4	2	8
2	4	6	4	5
3	4	3	1	5
2	3	5	10	3
7	10	1	8	1
8	3	2	4	2
3	2	10	9	1
7	2	10	2	6
6	6	1	6	6
5	10	5	1	4
8	10	3	2	8
4	2	3	6	3
9	5	7	4	6
6	1	1	6	2
10	6	9	3	9

(The numbers are read from left-to-right, then downwards.)

Using these values, I simulated what happens in the first 100 days in this modified definition of PEGG.

At the beginning, Y = "E", Z = 0.01, and K = 1000. The simulation starts as follows (See this post for the defintion of Letter Notation):

Date Days passed n X Letter Notation PEGG
2017-05-08 0 - 0 E0 1
2017-05-09 1 10 1 E1 10
2017-05-10 2 2 1.04 E1.04 10
2017-05-11 3 8 1.68 E1.68 47
2017-05-12 4 8 2.32 E2.32 208
2017-05-13 5 3 2.41 E2.41 257
2017-05-14 6 2 2.45 E2.45 281
2017-05-15 7 9 3.26 E3.26 1,819
2017-05-16 8 4 3.42 E3.42 2,630
2017-05-17 9 9 4.23 E4.23 16,982
2017-05-18 10 5 4.48 E4.48 30,199
2017-05-19 11 7 4.97 E4.97 93,325
2017-05-20 12 3 5.06 E5.06 114,815
2017-05-21 13 8 5.7 E5.7 501,187
2017-05-22 14 4 5.86 E5.86 724,435
2017-05-23 15 4 6.02 E6.02 1,047,128
2017-05-24 16 8 6.66 E6.66 4,570,881
2017-05-25 17 5 6.91 E6.91 8,128,305
2017-05-26 18 9 7.72 E7.72 52,480,746
2017-05-27 19 3 7.81 E7.81 64,565,422
2017-05-28 20 9 8.62 E8.62 416,869,383
2017-05-29 21 8 9.26 E9.26 1,819,700,858
2017-05-30 22 2 9.3 E9.3 1,995,262,314
2017-05-31 23 8 9.94 E9.94 8,709,635,899
2017-06-01 24 7 10.43 E10.43 26,915,348,039

At this point, we should switch to the letter F, and need to find a value m such that Fm = E10.43. Since rule (e-3) only requires us to reach a certain digits of precision after decimal point (currently 3 digits), we can use, say, binary search to determine the value.

Some calculations later, I found:

  • F2.007 = EE(10^0.007) = EE1.0162... = E10.3812...
  • F2.008 = EE(10^0.008) = EE1.0185... = E10.4373...

Therefore F2.007 < E10.43 < F2.008. So the new value for X is 2.007.

(Edit: Actually, you don't need binary search. Just reverse the calculation: E10.43 = EE(log(10.43)) = EE1.01828... = F(2+log(1.01828...)) = F2.00786...)

The values are updated to Y = "F", Z = 0.007, and K = 10000, and the simulation continues:

(In the table below, (10↑)A B means 10↑10↑...10↑B with A repetitions of "10↑")

Date Days passed n X Letter Notation PEGG
2017-06-01 24 - 2.007 F2.007 24,056,205,550
2017-06-02 25 2 2.035 F2.035 1,354,711,147,773
2017-06-03 26 5 2.21 F2.21 7.261863×1041
2017-06-04 27 5 2.385 F2.385 1.150188×10267
2017-06-05 28 4 2.497 F2.497 1.004780×101382
2017-06-06 29 2 2.525 F2.525 8.715435×102236
2017-06-07 30 8 2.973 F2.973 10↑2,495,934,049
2017-06-08 31 2 3.001 F3.001 10↑11,303,714,830
2017-06-09 32 4 3.113 F3.113 10↑10↑19.8234
2017-06-10 33 6 3.365 F3.365 10↑10↑207.6799
2017-06-11 34 4 3.477 F3.477 10↑10↑998.0734
2017-06-12 35 5 3.652 F3.652 10↑10↑30,722.3
2017-06-13 36 3 3.715 F3.715 10↑10↑154,170
2017-06-14 37 4 3.827 F3.827 10↑10↑5,179,508
2017-06-15 38 3 3.89 F3.89 10↑10↑57,872,356
2017-06-16 39 1 3.897 F3.897 10↑10↑77,375,091
2017-06-17 40 5 4.072 F4.072 10↑10↑1,402,139,082,982,998
2017-06-18 41 2 4.1 F4.1 (10↑)3 18.1520
2017-06-19 42 3 4.163 F4.163 (10↑)3 28.5403
2017-06-20 43 5 4.338 F4.338 (10↑)3 150.5600
2017-06-21 44 10 5.038 F5.038 (10↑)3 2,205,754,999,901
2017-06-22 45 3 5.101 F5.101 (10↑)4 18.2737
2017-06-23 46 7 5.444 F5.444 (10↑)4 602.1618
2017-06-24 47 10 6.144 F6.144 (10↑)5 24.7261
2017-06-25 48 1 6.151 F6.151 (10↑)5 26.0491
2017-06-26 49 8 6.599 F6.599 (10↑)5 9,373.79
2017-06-27 50 1 6.606 F6.606 (10↑)5 10,875.6
2017-06-28 51 8 7.054 F7.054 (10↑)5 36,676,929,804,362
2017-06-29 52 3 7.117 F7.117 (10↑)6 20.3789
2017-06-30 53 2 7.145 F7.145 (10↑)6 24.9096
2017-07-01 54 4 7.257 F7.257 (10↑)6 64.1466
2017-07-02 55 2 7.285 F7.285 (10↑)6 84.6301
2017-07-03 56 3 7.348 F7.348 (10↑)6 169.2135
2017-07-04 57 2 7.376 F7.376 (10↑)6 238.1443
2017-07-05 58 10 8.076 F8.076 (10↑)6 3,408,184,483,722,185
2017-07-06 59 9 8.643 F8.643 (10↑)7 24,855.1
2017-07-07 60 1 8.65 F8.65 (10↑)7 29,297.8
2017-07-08 61 7 8.993 F8.993 (10↑)7 6,920,079,082
2017-07-09 62 2 9.021 F9.021 (10↑)7 161,573,158,948
2017-07-10 63 10 9.721 F9.721 (10↑)8 182,042
2017-07-11 64 2 9.749 F9.749 (10↑)8 407,830
2017-07-12 65 6 10.001 F10.001 (10↑)8 11,303,714,830

(Notice that the simulated PEGG first surpasses a googol on June 4, and a googolplex on June 10.)

At this point, another switch is required. Since 10.001 is very close to 10, we expect that the new value of X is 2. In fact:

  • G2.0000 = FF1 = F10
  • G2.0001 = FF(10^0.0001) = FF1.00023... = FE(10^0.00023...) = FE1.00053... = F10.01222...

Therefore G2.0000 < F10.001 < G2.0001, and the new value for X is 2.

(Edit: Using the reverse-calculation method, F10.001 = FE1.000043... = FF1.000018... = G2.0000081...)

The values are updated to Y = "G", Z = 0.0049, and K = 100000, and the simulation continues:

(In the table below, (10↑↑)A B means 10↑↑10↑↑...10↑↑B with A repetitions of "10↑↑")

Date Days passed n X Letter Notation PEGG
2017-07-12 65 - 2 G2 (10↑)8 10,000,000,000
2017-07-13 66 6 2.1764 G2.1764 (10↑)1478 92.9788
2017-07-14 67 6 2.3528 G2.3528 10↑↑(7.306981×1061)
2017-07-15 68 1 2.3577 G2.3577 10↑↑(2.778914×1079)
2017-07-16 69 6 2.5341 G2.5341 10↑↑10↑10↑430.3233
2017-07-17 70 6 2.7105 G2.7105 10↑↑(10↑)4 23.0717
2017-07-18 71 5 2.833 G2.833 10↑↑(10↑)5 2,644,504
2017-07-19 72 10 3.323 G3.323 10↑↑10↑↑(4.146693×1018)
2017-07-20 73 5 3.4455 G3.4455 10↑↑10↑↑10↑1,433,686
2017-07-21 74 1 3.4504 G3.4504 10↑↑10↑↑10↑4,186,579
2017-07-22 75 4 3.5288 G3.5288 10↑↑10↑↑10↑10↑247.6407
2017-07-23 76 8 3.8424 G3.8424 10↑↑10↑↑(10↑)5 1,121,962,254
2017-07-24 77 10 4.3324 G4.3324 (10↑↑)3 (6.572838×1025)
2017-07-25 78 3 4.3765 G4.3765 (10↑↑)3 (1.496869×10249)
2017-07-26 79 2 4.3961 G4.3961 (10↑↑)3 (4.772795×101219)
2017-07-27 80 8 4.7097 G4.7097 (10↑↑)3 (10↑)4 21.5647
2017-07-28 81 4 4.7881 G4.7881 (10↑↑)3 (10↑)5 23.8405
2017-07-29 82 2 4.8077 G4.8077 (10↑↑)3 (10↑)5 441.6540
2017-07-30 83 3 4.8518 G4.8518 (10↑↑)3 (10↑)6 19.2696
2017-07-31 84 6 5.0282 G5.0282 (10↑↑)3 (10↑)13 80,623.3
2017-08-01 85 3 5.0723 G5.0723 (10↑↑)3 (10↑)31 259,926,582
2017-08-02 86 9 5.4692 G5.4692 (10↑↑)4 10↑670,322,966
2017-08-03 87 5 5.5917 G5.5917 (10↑↑)4 10↑10↑111,791,645
2017-08-04 88 7 5.8318 G5.8318 (10↑↑)4 (10↑)5 1,414,108
2017-08-05 89 4 5.9102 G5.9102 (10↑↑)4 (10↑)7 22.6643
2017-08-06 90 6 6.0866 G6.0866 (10↑↑)4 (10↑)44 457,650,025
2017-08-07 91 6 6.263 G6.263 (10↑↑)5 6,265,495
2017-08-08 92 1 6.2679 G6.2679 (10↑↑)5 13,497,459
2017-08-09 93 1 6.2728 G6.2728 (10↑↑)5 30,475,879
2017-08-10 94 6 6.4492 G6.4492 (10↑↑)5 10↑3,193,246
2017-08-11 95 2 6.4688 G6.4688 (10↑↑)5 10↑590,664,525
2017-08-12 96 10 6.9588 G6.9588 (10↑↑)5 (10↑)8 17.5530
2017-08-13 97 6 7.1352 G7.1352 (10↑↑)5 (10↑)207 46.0697
2017-08-14 98 9 7.5321 G7.5321 (10↑↑)6 10↑10↑346.8867
2017-08-15 99 3 7.5762 G7.5762 (10↑↑)6 10↑10↑744,433
2017-08-16 100 9 7.9731 G7.9731 (10↑↑)6 (10↑)8 322.4052

This concludes the first 100 days of the simulation of (modified) PEGG. I think the real PEGG would behave similarly, provided that the "Dow Johns rule" indeed produces pseudorandom numbers.

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