Official "Benford was wrong!" thread

Mini-Geek · 2009-06-16, 19:42

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:

and odd first, then even:

Code:

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?

10metreh · 2009-06-17, 06:20

Quote:

Originally Posted by Mini-Geek

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

CRGreathouse · 2009-06-17, 06:49

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

Batalov · 2009-06-17, 07:13

Quote:

Originally Posted by Mini-Geek

...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!

10metreh · 2009-06-17, 07:21

Quote:

Originally Posted by Batalov

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?

CRGreathouse · 2009-06-17, 07:39

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.

Greebley · 2009-06-17, 13:52

Quote:

Originally Posted by Batalov

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?

10metreh · 2009-06-17, 14:16

Quote:

Originally Posted by Greebley

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?

Greebley · 2009-06-18, 15:23

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.

10metreh · 2009-06-18, 15:29

Quote:

Originally Posted by Greebley

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.

Greebley · 2009-06-18, 15:54

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.

2009-06-16, 19:42	#1
Mini-Geek Account Deleted "Tim Sorbera" Aug 2006 San Antonio, TX USA 7·13·47 Posts	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?

2009-06-17, 06:49	#3
CRGreathouse Aug 2006 3·1,993 Posts	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

2009-06-17, 07:39	#6
CRGreathouse Aug 2006 3·1,993 Posts	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. Attached Thumbnails

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2009-06-18, 15:23	#9
Greebley May 2009 Dedham Massachusetts USA 34B₁₆ Posts	3^45^27^2 = 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.

2009-06-18, 15:54	#11
Greebley May 2009 Dedham Massachusetts USA 1513₈ Posts	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 2p0^2 => p0^2 + 3P + 1 (odd) = p1q1 => p2q2 => p3q3... => 99225. Note that p0 could be arbitrarily large because there is an arbitrarily large p1q1 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 p1q1 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.