20150615, 22:52  #1 
"Gang aft agley"
Sep 2002
3,581 Posts 
Is every product of distinct Fermat numbers symmetrical in binary?
Quote by David Richeson:
Here's what I learned today: Write out the mod2 version of Pascal's triangle. That is, write out Pascal's triangle as usual and replace each even number with 0 and each odd number with 1. Then view each row as a binary number. This list of numbers is the full list of all products of distinct Fermat numbers. As we see in the figure the first several rows are 1, 3, 5, 15=3*5, 17, 51=3*17, 85=5*17, 255=3*5*17, 257,... 
20150616, 10:31  #2 
"Matthew Anderson"
Dec 2010
Oregon, USA
26E_{16} Posts 
Every Fermat number is symmetric in Binary. F(n) = 2^(2^n)+1. The first digit is a 1 then there are some zeros and finally another 1. F(1) = 5 = 101 in base 2. F(2) = 17 = 10001 in base 2. Regards, Matt 
20150616, 11:03  #3 
Jun 2003
17·281 Posts 
Yes. We can prove this using induction. Note that product(n=0,N1, F(n)) = =M(2^n) = F(N)2

20150616, 11:22  #4  
"Gang aft agley"
Sep 2002
3,581 Posts 
I know that every Fermat number is symetrical. What surprised me is that the product of multiple distinct Fermat numbers is also symetrical in binary.
Why did this surprise me? Because although only two bits are set in each Fermat number, when multiplying out a product of multiple distinct Fermat numbers, it seems that that there will be some carries to the left, but this result is still symetrical. The statement that every possible product of distinct Fermat numbers is symetrical is implicit in the symmetry of Pascal's Table in my spoiler quoted section. Take a line of the table, perform a mod 2, on each entry, and the remaining even/odd bits of each entry grouped together as a binary string results is a symetrical binary number for that row and this number is also a product of distinct Fermat numbers. I do admit that when something surprises me it usually means that I didn't understand it and that is why I immediately gave a quote that showed the implication that I titled the thread upon, so that even if I was off in the weeds somehow, I could intend to avert significant time expenditures but still present a puzzle. The thread title was intended to be an exception to Betteridge's law of headlines. Quote:
Last fiddled with by only_human on 20150616 at 12:15 Reason: s/advert/avert/ s/parity/even//odd/ 

20150617, 17:49  #5 
May 2004
New York City
4198_{10} Posts 
IS THAT SO?

20150617, 17:54  #6 
"Gang aft agley"
Sep 2002
3,581 Posts 

20150618, 17:55  #7 
"William"
May 2003
New Haven
2^{2}·3^{2}·5·13 Posts 
Isn't this the faulty assumption? Multiplying them sequentially from smallest to largest, at each step the larger is so much larger that the two images of the smaller (using long hand "school" multiplication) never overlap, so there are never any carries.
Last fiddled with by wblipp on 20150618 at 17:55 
20150618, 18:12  #8  
"Gang aft agley"
Sep 2002
3,581 Posts 
Quote:
Perhaps it's not much of a puzzle but I think it is kind of pretty. The numbers remind of fractals or Gray Code tables in that the larger entries contain the earlier entries as substrings. Last fiddled with by only_human on 20150618 at 18:32 Reason: forgot "not" "occur", deleted "all" as possibly ambiguous in "all the earlier", s/Grey/Gray/ (Frank Gray) 

20150626, 09:35  #9  
"Gang aft agley"
Sep 2002
3,581 Posts 
Quote:
So how does one generate this list of all products of distinct Fermat Numbers? Of all Gray Code tables, in my opinion, the most convenient Gray Code is based on k = k XOR ( k >> 1 ) This keeps the most significant bit unchanged and the bit shifted off to the right that gets neglected during the calculation is unneeded for that result. But if you wanted an unchanged bit on both sides (and a wider result), it is easier to shift to the left like this: k = k XOR ( k << 1 ) Why does this matter here? Because starting with k=1 and applying this line repeatedly generates the the exact values that are also generated from a Mod 2 operation on a normal Pascal's table: in binary: 1 This generates the numbers that I was looking at for this thread. As you can see, they are symmetrical and the above quoted reference has a link to a proof by induction that these numbers are a full list of products of distinct Fermat numbers:11 101 1111 10001 110011 1010101 ... Quote:


20150626, 10:30  #10 
May 2004
New York City
1000001100110_{2} Posts 
If Fn = 2^2^n+1, then F0 = 2^2^0+1 = 3, F1 = 5, F2 = 17, F3 = 257, etc.
Then Fn = F0 * F1 * F2 * ... * Fn1 + 2 for n >= 1. So F0 * F1 * F2 * ... * Fn1 is "all ones" in binary. Since the "number of bits  1" is 1, 2, 4, 8, 16, ... , 2^n for Fn, their products of distinct Fn's never overlap in "bitspace." There are no carries, so all such products remain symmetric. After F4 = 65537, no higher Fn is known to be prime. Their factors are not in general symmetric. (I just noticed axn already said some of this. Hope the redundancy helps.) Last fiddled with by davar55 on 20150626 at 10:32 Reason: credit axn 
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