What's the Next Term II

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Well, I don’t think I’m going to suck in anyone who hasn’t been heard from by now…

So, OK, what is the next term of this sequence? 0, 1, 2, 8.252 x 10^2466, ...

Here, "10^2466" means "ten to the 2466th power." The corresponding term is necessarily slightly approximate, since it contains 2467 decimal digits.

Here’s my answer (the one that the problem comes with): t(n) = n(!^n).

0 = 0
1! = 1
2!! = 2! = 2
3!!! = 6!! = 720! = 8.252 x 10^2466
4!!!! is a rather large number in most contexts

Hey, you wanna see a really big number? Howsabout (10^10)(!^(10^10))? True, it’s still only 0% of aleph-null.

Comments

  1. gcallah7:51 AM

    "4!!!! is a rather large number in most contexts."

    But not that of, "Shots of well vodka consumed by Wabulon at B61.

    ReplyDelete
  2. LarryRuane12:21 PM

    Are you sure about 3!!!? Using linux bc -l funcs.bc, both "factorial(720)" and "factorial(factorial(factorial(3)))" produce:

    26012189435657951002049032270810436111915218750169457857275418378508\
    35631156947382240678577958130457082619920575892247259536641565162052\
    01587379198458774083252910524469038881188412376434119195104550534665\
    86162432719401971139098455367272785370993456298555867193697740700037\
    00430783758997420676784016967207846280629229032107161669867260548988\
    44551425719398549944893959449606404513236214026598619307324936977047\
    76060676806701764916694030348199618814556251955925669188308255149429\
    47596537274845624628824234526597789737740896466553992435928786212515\
    96748322097602950569669992728467056374713753301924831358707612541268\
    34158601294475660114554207495899525635430682886346310849656506827715\
    52996256790845235702552186222358130016700834523443236821935793184701\
    95651072978180435417389056072742804858399591972902172661229129842051\
    60675790362323376994539641914751755675576953922338030568253085999774\
    41675784352815913461340394604901269542028838347101363733824484506660\
    09334848444071193129253769465735433737572477223018153403264717753198\
    45373414786743270484579837866187032574059389242157096959946305575210\
    63203263493209220738320923356309923267504401701760572026010829288042\
    33560664308988871029738079757801305604957634283868305719066220529117\
    48225105366977566030295740433879834715185526028053338663571391010463\
    36419769097397432285994219837046979109956303389604675889865795711176\
    56667003915674815311594398004362539939973120306649060132531130471902\
    88984918562037666691644687911252491937544258458950003115616829743046\
    41142538074897281723375955380661719801404677935614793635266265683339\
    50976000000000000000000000000000000000000000000000000000000000000000\
    00000000000000000000000000000000000000000000000000000000000000000000\
    00000000000000000000000000000000000000000000000

    which has around 1746 digits, not 2466. Great puzzle though; thanks for posting it.

    ReplyDelete
  3. gcallah12:38 PM

    I get the same answer on Max OS X bc -- which, of course, is the same program, so no surprise. The question I have is, "Do we have reason to think bc is reliable for such huge numbers?"

    (Of course, another question is, "Do we have reason to think Wabulon is reliable for such huge numbers?")

    ReplyDelete
  4. gcallah12:42 PM

    Larry, is there a way to force bc to output in scientific notation?

    ReplyDelete
  5. LarryRuane1:26 PM

    Yes, funcs.bc provides it (and at least on Linux that's what implements factorial(), so I assume you have it). For example:

    engnot(factorial(720), 5)
    2.60121*10^1746

    ReplyDelete
  6. Wabulon2:12 PM

    Err, uhh, I used Stirling's Formula (quite accurate for approximating large factorials), log10-ing everything. And, no, I am not particularly good at arithmetic (although I have a nice essay on Peano's Postulates if anyone wants to see it). And, of course, I really don't much care what the actual value of 720! is, it's the principle of the thing...

    ReplyDelete
  7. gcallah3:59 PM

    Yes, but Wabulon, how are we supposed to guess the formula if one of the items in the series is wrong?

    ReplyDelete
  8. Wabulon4:21 PM

    Larry -

    I used Stirling's formula in the form: log n! approximately =
    (n+0.5) log n - 0.4343n + 0.3991.
    Those coefficients checked out when I recalculated them, but, obviously, my arithmetic didn't, cuz when I redid it for 720!, I got roughly 2x10^1746. Sorry.

    ReplyDelete

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