A129755 (??!!)

[Andrew Plewe]

On 6/1/07, Andrew Plewe <aplewe at sbcglobal.net> wrote:

> There is one property of constant-spaced integers which allows for the easy
> factorization of some of them. Consider the following representation of 35:

Roughly speaking, one representation gives one factor of k.
For k=35, it is enough since 35 is the product of just two primes. But
in general, knowing one factor of a number in not enough to complete
its factorization.

Max




If you really want to enter the *Special* Olympics, there are several 

* Take any existing sequence that contains 185 and artificially 
offset it to get a new one that starts with 185, say:


* Make up some expression such that a(1) = 185 based on some random 
representation of 185:


Arbitrarily replacing a few 1's with n's we can obtain the 
mysteriously impressive submission:

110308805, 1779292086, 28990080007, ...

Or... (inspired by a Four-4's representation):



where the next term is ~197 digits.  (Not sure all a(n) are 
integers?  Just submit "Numerator of...").

Or... view that 3,4,5 as sides a,b,a^2+b^2 of the "first" Pythagorean 
triangle, and then index through the rest of the a,b...

* Take a sequence such that b(K)=185 for some K, find a sequence 
whose smallest term c(1)=K, then submit a(n)=b(c(n)).

* From the existing sequences containing 185, intersect groups of 
them to find a combo where the smallest term is 185.  Then submit 
"Numbers that are blah and blah and...".

And so on.

But, arguably, without additional motivation, NONE of these hack 

Right?





Good point, I should clarify that within the given limits this technique can
be used to find one or two non-obvious factors of k. I say two because if
the length of a representation is odd, then the middle value will divide k,
thus you can get two factors if the length is not equal to the middle value.
For instance:

5+7+9+11+13+15+17 = 77, middle value = 11, 77/11 = 7

The technique I demonstrated in my last email can be converted into an
algorithm, here's the short version of how it works for triangle numbers.
First, find the representation k = tri(x) + r, where r < x. For the second
example I gave, 68, we have tri(x) + r = 68 = tri(11) + 2. Essentially I
want to convert tri(x) + r into the form tri(y) + yz. This is possible if
the following quadratic has at least one positive integer solution (I leave
out the derivation because my version is somewhat long):

1.) (-3 +/1 sqrt(9 + 8(x - r))) / 2

x = 11, r = 2, therefore x - r = 9 and we have (-3 +/- sqrt(9 + 8(9))) / 2 =
(-3 +/- sqrt(81)) / 2 = 6/2 = 3, and -12/2 = -6. I can now find y by
tri(8) + (8)z = 68. GCD(k, y) will immediately produce one factor of 68; 4.
If y were odd, I could continue to find the "middle" value of this
particular representation of k. In this case y = 8 so there is no "middle"
value. The first value in the representation will be z + 1. 68 - tri(8) =
32, thus (8)z = 32 and z = 4. So our starting value is z + 1 = 4 + 1 = 5,
and the full representation is:

5 6 7 8 9 10 11 12 = tri(8) + (8)4 = 68

The values of (x - r) for which equation 1 has integer solutions are
A000096. Equation 1 can be generalized for all sums of constant-spaced
integers, I don't remember how off the top of my head. The trick, of course,
is to know which generalized version of equation 1 will work for an integer
k of unknown factorization. The bigger question is whether or not such a
quadratic exists for all composite k.



-----Original Message-----

On 6/1/07, Andrew Plewe <aplewe at sbcglobal.net> wrote:

> There is one property of constant-spaced integers which allows for the
easy
> factorization of some of them. Consider the following representation of
35:

Roughly speaking, one representation gives one factor of k.
For k=35, it is enough since 35 is the product of just two primes. But
in general, knowing one factor of a number in not enough to complete
its factorization.

Max