Playing with numbers...: zorac

Playing with numbers...

I was idly playing with numbers in my head the other day, something to occupy my mind while I trudged back from Sainsbury's with a load of shopping. In particular, I was looking at number sequences (you know, what number comes next - 1, 3, 5, 7, ? or 1, 4, 9, 16, ?) and came up with the following (related, but calculated in opposite ways):

1 ... 4 ... 19,683 ... ?
1 ... 4 ... 7,625,597,484,987 ... ?

The answer to the first series is a mere
340,282,366,920,938,463,463,374,607,431,768,211,456
The second, as you can probably guess, is larger. Much larger. In fact, it's exactly
2^{26,815,615,859,885,194,199,148,049,996,411,692,254,958,731,641,184,786,755,447,122,887,443,528,060,147,093,953,603,748,596,333,806,855,380,063,716,372,972,101,707,507,765,623,893,139,892,867,298,012,168,192}
or approximately
5 x 10^{8,072,304,726,028,225,379,282,369,632,412,842,737,810,037,345,818,396,835,407,833,195,922,269,447,488,939,454,873,506,883,657,604,175,736,107,203,603,983,535,115,431,016,379,958,411,831,793,261,615,652,260}
Even more approximately, that's
5 x 10^{8 x 10¹⁵³}
Time to play with some big numbers:
5 x 10^{googol x 8 x 10⁵³}
5 x googolplex^{8,000 x 10⁵⁰}
5 x googolplex^{8,000 √googol}

However I put it, that is one very big number, almost impossible to visualise - Asimov's essay Skewered! details just how hard to visualise big numbers can be. Final challenge, if you want to prove that you actually solved the puzzle: explain how the series was derived.

jiggery_pokery

September 5 2003, 13:44:26 UTC 17 years ago

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I'm guessing at - let's see if HTML is good enough to be able to produce this -

1 , 2², 3^3³, 4^{4^4⁴}

and a similar thing for the second one but instead of there being n recursions of "powers of n", there are n! recursions of "powers of n".

Yay HTML! :-)

Deleted comment

zorac

September 5 2003, 14:16:15 UTC 17 years ago

I can't argue with that. Trying to calculate those numbers in one's head is however difficult. Particularly when they come out differently in different directions - as explained below, both of the series are the same, just with the parentheses in different places. Using smaller numbers (and no more sups)

((2^2)^2)^2 = (4^2)^2 = 16^2 = 256
2^(2^(2^2)) = 2^(2^4) = 2^16 = 65,536

With 4s in place of the 2s, you get the original final answers.

imc

September 5 2003, 14:51:27 UTC 17 years ago

But of course ((4^4)^4)^4 is the same as 4^(4*4*4) so your first series condenses to that of n^(n^(n-1)).

September 5 2003, 16:08:50 UTC 17 years ago

Ooh! That's neat. Should've figured that one myself.

Unfortunately, the second series isn't no neatly condensing (unless there's an operater which is to ^ as Σ is to + and (IIRC) Π is to *). Best I can do is functional:

f(n) = g(n, n);
g(x, 0) = 1;
g(x, y + 1) = x ^ g(x, y)

17 years ago

September 7 2003, 11:08:19 UTC 17 years ago

^{It's^{my^{journal^{I^{can^{do^{what^{I^want!}}}}}}}}

September 5 2003, 13:58:28 UTC 17 years ago

Playing about a bit, I think that the second one is what we both identified for the first (1, 2², 3^3³, 4^{4^4⁴}) and that the first one is the slightly more sane 1¹, 2^2¹, 3^{3^2¹}, 4^{4^{3^2¹}}.

Go go Google Calculator!

September 5 2003, 14:00:20 UTC 17 years ago

...or maybe not.

September 5 2003, 15:03:08 UTC 17 years ago

Is it a bird?
Is it a plane?
No! It's <sup>-erman!

September 5 2003, 14:00:11 UTC 17 years ago

And in a mere 6 minutes from my original posting ;-)
Actually, they're poth the series you gave, just calculated in oposite directions, the first being
1, 2², (3³)³, ((4⁴)⁴)⁴
and the second
1, 2², 3^(3³), 4^{(4^(4⁴))}