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Old 2009-06-16, 19:42   #1
Mini-Geek
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Default Official "Benford was wrong!" thread

optional full title: Mini-geek's Mindless, Most likely Meaningless, Mundane Mathematical Musings (with Much alliteration!)


10metreh: All right, I've added one more word, but the title box isn't long enough for all of them

After noticing a large number of factors in Aliquot sequences start with 1, I decided to count it. 134856 seemed like a good candidate considering its length, so I decided to count this sequence's factors. Note that the smallest factor is always excluded in these counts. (it was 2 or a power of 2 in all lines except the last one, which is 3709 lines...is this a mathematical requirement or just a very rare occurrence?) Anyway, here are the counts, sorted by the digit: (for the curious: there are 16091 non-smallest factors)
Code:
1 4787 29.75%
2 2237 13.90%
3 2841 17.66%
4 1291  8.02%
5 1850 11.50%
6  782  4.86%
7 1205  7.49%
8  590  3.67%
9  508  3.16%
sorted by the number of occurrences:
Code:
1 4787
3 2841
2 2237
5 1850
4 1291
7 1205
6  782
8  590
9  508
and odd first, then even:
Code:
1 4787
3 2841
5 1850
7 1205
9  508
2 2237
4 1291
6  782
8  590
I'm sure that due to Benford's Law we should expect a tendency towards lower numbers, but I wouldn't think it should be anywhere near this big of a difference. (I think the even/odd difference can be attributed to the divisibility of single-digit factors) Anyone know if statistically this is just normal or if there's anything else going on?

Last fiddled with by 10metreh on 2009-06-17 at 10:54 Reason: Why not?
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Old 2009-06-17, 06:20   #2
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Quote:
Originally Posted by Mini-Geek View Post
I'm sure that due to Benford's Law we should expect a tendency towards lower numbers, but I wouldn't think it should be anywhere near this big of a difference. (I think the even/odd difference can be attributed to the divisibility of single-digit factors) Anyone know if statistically this is just normal or if there's anything else going on?
I think it is that big
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Old 2009-06-17, 06:49   #3
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Here's roughly what you'd expect from Benford's Law and single-digit divisors.
Code:
n      true    expected
1	4787	4089
2	2237	2392
3	2841	2933
4	1291	1316
5	1850	1817
6	782	909
7	1205	1318
8	590	695
9	508	622
That seems to track reasonably well.

Edit: Here's roughly what you'd expect from Benford's Law and divisors below 101.
Code:
n      true    expected
1	4787	4425
2	2237	2272
3	2841	2863
4	1291	1347
5	1850	1766
6	782	870
7	1205	1333
8	590	662
9	508	553

Last fiddled with by CRGreathouse on 2009-06-17 at 07:11
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Old 2009-06-17, 07:13   #4
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Quote:
Originally Posted by Mini-Geek View Post
...it was 2 or a power of 2 in all lines except the last one, which is 3709 lines...is this a mathematical requirement or just a very rare occurrence?
Odd number life-cycle:
- if prime, it's the end
- else if no even-powered odd primes, will stay odd
- else (e.g. with just 32 or 52, which is far from improbable) may become even
- else drops very fast (therefore, all starting odd values are removed long ago)

Even number life-cycle:
- will stay even, except being a square or 2*square (all even-powered odd primes with any power of 2*)

This is basically why.

____

*silly me, overlooked the specialness of 2!

Last fiddled with by Batalov on 2009-06-17 at 08:10 Reason: (2 is special)
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Old 2009-06-17, 07:21   #5
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Quote:
Originally Posted by Batalov View Post
Even number life-cycle:
- will stay even, except being a square (all even-powered primes)
Doesn't 2 * an odd square do it as well?
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Old 2009-06-17, 07:39   #6
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Here's a graph of the true counts and those using Benford's law and primes up to 10^0 / 10^1 / 10^2. You can see that once you include even the one-digit primes the prediction is very close to the true values.

Heh, apologies for the cursor.
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Old 2009-06-17, 13:52   #7
Greebley
 
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Quote:
Originally Posted by Batalov View Post
Odd number life-cycle:
- if prime, it's the end
- else if no even-powered odd primes, will stay odd
- else (e.g. with just 32 or 52, which is far from improbable) may become even
- else drops very fast (therefore, all starting odd values are removed long ago)
I think to go from odd to even it has to be a perfect square. Any odd prime to an odd power is even which makes the product even which will keep the result odd

As an aside, I was trying to guess what the smallest odd number to reach 100 digits is and my guess is 5^6 = 15625 which goes up to 140 digits. Anyone know a smaller one?

Last fiddled with by Greebley on 2009-06-17 at 13:52
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Old 2009-06-17, 14:16   #8
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Quote:
Originally Posted by Greebley View Post
As an aside, I was trying to guess what the smallest odd number to reach 100 digits is and my guess is 5^6 = 15625 which goes up to 140 digits. Anyone know a smaller one?
Another one: What is the smallest odd number to reach 100 digits without becoming a side-sequence?
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Old 2009-06-18, 15:23   #9
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3^4*5^2*7^2 = 99225 is certainly a candidate but it isn't in the list. It ends at 72 digits in the db however so I am unsure if 100 is even reached.

Since it isn't in the list, but is less than 100k, I am guessing something merges with it or it decreases again and is not the answer.

Note that proving that a sequence isn't a side sequence is beyond our current abilities as far as I know.

I may try running it it a bit and see what it does.
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Old 2009-06-18, 15:29   #10
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Quote:
Originally Posted by Greebley View Post
Note that proving that a sequence isn't a side sequence is beyond our current abilities as far as I know.
It is very hard...

...but there are databases on Wolfgang's site which have a very high chance of finding a merge if one exists.
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Old 2009-06-18, 15:54   #11
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Here is my argument that it can't be done:

There is a conjecture that has a good chance of being true that every even number is the sum of two primes. Therefore for every odd number sufficiently large there is a sum of p + q + 1 for two odd primes p and q. But for the aliquot sequence pq => p+q+1. This means that one can find arbitrarily large products of two primes that lead to that odd number. Now say 276 hit 2*p0^2 => p0^2 + 3P + 1 (odd) = p1*q1 => p2*q2 => p3*q3... => 99225. Note that p0 could be arbitrarily large because there is an arbitrarily large p1*q1 that will lead to 99225.

So one has to complete every sequence less than 99225 (or whatever the real answer is) to actually prove it. You might be able to argue that the chance of 276 hitting a p1*q1 that leads to 99225 is very unlikely, but you do have an infinite number of chances if 276 went to infinity (the chance can still sum to much less than 1 but you would have to show that to show it was unlikely.
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