20090616, 19:42  #1 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
7·13·47 Posts 
Official "Benford was wrong!" thread
optional full title: Minigeek'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 nonsmallest 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% Code:
1 4787 3 2841 2 2237 5 1850 4 1291 7 1205 6 782 8 590 9 508 Code:
1 4787 3 2841 5 1850 7 1205 9 508 2 2237 4 1291 6 782 8 590 Last fiddled with by 10metreh on 20090617 at 10:54 Reason: Why not? 
20090617, 06:20  #2  
Nov 2008
2×3^{3}×43 Posts 
Quote:


20090617, 06:49  #3 
Aug 2006
3·1,993 Posts 
Here's roughly what you'd expect from Benford's Law and singledigit 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 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 20090617 at 07:11 
20090617, 07:13  #4  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
11×19×47 Posts 
Quote:
 if prime, it's the end  else if no evenpowered odd primes, will stay odd  else (e.g. with just 3^{2} or 5^{2}, which is far from improbable) may become even  else drops very fast (therefore, all starting odd values are removed long ago) Even number lifecycle:  will stay even, except being a square or 2*square (all evenpowered odd primes with any power of 2*) This is basically why. ____ *silly me, overlooked the specialness of 2! Last fiddled with by Batalov on 20090617 at 08:10 Reason: (2 is special) 

20090617, 07:21  #5 
Nov 2008
2×3^{3}×43 Posts 

20090617, 07:39  #6 
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 onedigit primes the prediction is very close to the true values.
Heh, apologies for the cursor. 
20090617, 13:52  #7  
May 2009
Dedham Massachusetts USA
3×281 Posts 
Quote:
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 20090617 at 13:52 

20090617, 14:16  #8 
Nov 2008
2·3^{3}·43 Posts 
Another one: What is the smallest odd number to reach 100 digits without becoming a sidesequence?

20090618, 15:23  #9 
May 2009
Dedham Massachusetts USA
34B_{16} Posts 
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. 
20090618, 15:29  #10 
Nov 2008
2·3^{3}·43 Posts 

20090618, 15:54  #11 
May 2009
Dedham Massachusetts USA
1513_{8} 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 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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