| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I can't remember if this was discussed in here yet or not. In the brain_boggler's conference last year I pozed the puzzler of how to produce the number 4 from just 3 and factorial and integer square root. Here's how to get 1,2,3, and 5: 1 = sqrt(3) (sqrt means [sqrt] means integer sqrt) 2 = sqrt(3!) 3 = 3 5 = sqrt(sqrt(3!!)) 4 is harder to get, but someone finally answered in that notes file. I think the original rec.puzzles article said Knuth claimed all integers could be reached from 3, through proper application of integer sqrt and factorial. Can any of you readers explain the proof of this ? If not, can you think of any ways to explore this on the computer ? Straight-forward means fail miserably due to the huge size that factorials reach. For instance, is there any way to evaluate things like: sqrt(sqrt(sqrt(3!!!!))) without having to figure out all those extra digits of 3!!!!. It seems a shame to have to generate them if they're just going to be thrown away again when we take integer square root. Hmmm, I just thought of something: 3! = 3 * 2 * 1 = 6 3!! = 6! = 6 * 5 * 4 * 3 * 2 * 1 = 2^4 * 3^2 * 5 (writing 3!! as product of powers of primes) 3!!! = (2^4 * 3^2 * 5)! = ??? Any way we can get 3!!! as product of powers of primes without knowing that it's 720! ? If so, then we could come up with a way to represent any n!!!...!! as a product of powers of primes. Such a representation can easily be sqrted by just halving all the even exponents (and doing I don't know what with the odd ones :-) Anyone want to take on from here, or is this just pure mathematical rubbish ? /Eric
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1370.1 | GUESS::DERAMO | Dan D'Eramo | Sun Jan 13 1991 18:16 | 27 | |
We start with 3.
Take isqrt and factorial of that and you add 1 and 6.
Add in isqrt and factorial of those and you get 1, 1, 2,
and 720. We already have 1, so the new results were 2
and 720.
The next step adds 1, 2, 26, and 720!
From here I will ignore the big numbers and the numbers
already considered. With that, the next steps reach:
26 -> 5, 26!
5 -> 2, 120
120 -> 10, 120!
10 -> 3, 3628800
10! -> 1904, (10!)!
1904 -> 43, 1904!
43 -> 6, 43!
6 was reached already. So we have reached and continued
from 1, 2, 3, 5, 6, 10, 26, 43, 120, 720, 1904, and 10!;
and reached but not yet continued from 26!, 43!, 120!,
720!, 1904!, and 3628800!. Time to call in Captain Bignum :-)
Dan
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| 1370.2 | GUESS::DERAMO | Dan D'Eramo | Sun Jan 13 1991 23:54 | 16 | |
4 was reached on the twentieth iteration. You showed how
to reach 5; 5! is 120; seven isqrt's of 120! is 35; six
isqrt's of 35! is 4. Using s for integer sqrt and f for
factorial the sequence from left to right is
3ffssffsssssssfssssss is 4
or in LISP, (isqrt (isqrt (isqrt (isqrt (isqrt (isqrt
(factorial (isqrt (isqrt (isqrt (isqrt (isqrt (isqrt
(isqrt (factorial (factorial (isqrt (isqrt (factorial
(factorial 3))))))))))))))))))))
There might be a shorter way; I did not check factorials
of anything greater than 999.
Dan
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| 1370.3 | helpful direction suggester | HERON::BUCHANAN | Holdfast is the only dog, my duck. | Mon Jan 14 1991 10:52 | 7 |
I'm a long way from my desk here. Can someone retrieve Stirling's formula for the limiting behaviour of n!, together with its error term (which I seem to recall is O(n/12)). Maybe we can then say something more solid about which n are achievable by Eric's rather strange mechanism. Cheers, Andrew. | |||||
| 1370.4 | ALLVAX::JROTH | Saturday alley up to Sunday street | Mon Jan 14 1991 11:23 | 6 | |
gamma(z) ~ exp(-z) z^(z-1/2) sqrt(2 pi) *
[ 1 + 1/(12z) + 1/(288 z^2) - 139/(51840 z^3) - 571 / (24888320 z^4) ... ]
- Jim
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| 1370.5 | GUESS::DERAMO | Dan D'Eramo | Mon Jan 14 1991 12:43 | 3 | |
And n! = gamma(n+1).
Dan
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| 1370.6 | Another Hailstone? | CIVAGE::LYNN | Lynn Yarbrough @WNP DTN 427-5663 | Mon Jan 14 1991 14:48 | 4 |
It looks offhand as though this problem is a variant of the Hailstone Problem, discussed elsewhere, and about as intractable. It will be interesting to see if this problem can be attacked by any methods that might then be applied to the other. | |||||
| 1370.7 | GUESS::DERAMO | Dan D'Eramo | Mon Jan 14 1991 19:05 | 8 | |
re .-1,
>> and about as intractable.
Actually, the base note echoes a claim that this problem
has been solved.
Dan
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