## Re: Q:Certain group of composites and their prime factors.

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• ... Are consecutive solutions to a quadratic equation f(1)=11, f(2)=13,... ... For any solution f(a)=p*q, there will exist the solution sequences f
Message 1 of 2 , Aug 30, 2004
--- In primenumbers@yahoogroups.com, "Dan" <30pack@s...> wrote:
>
> Primes of this form --
>
> 11 +2,+4,+6,+8,+10,+12,+14,..+n

Are consecutive solutions to a quadratic equation f(1)=11, f(2)=13,...

>
> Where if at any point the sum is a composite then these composites
> will not have any of these primes in their prime factors.

For any solution f(a)=p*q, there will exist the solution sequences f
(a+k*q)==0(modq) and f(a+k*p)==0(modp) for all integer k. By
quadratic reciprocity, there will also be a second set of solutions f
(a+k*q+b)==0(modq) and f(a+k*p+c)==0(modp). Generally, either no
solution sequence exists, or two distinct sequences exist. The only
way a single sequence occurs is if the factor in question divides the

In your example above, with 2nd derivative=2, all factors have two
solution sequences. The quadratic progression is symmetrical in that
f(0)=f(1), f(-1)=f(2),f(-2)=f(3),... Thus, any prime factor, q,
appearing in the progression at f(a), where a is the lowest positive
index for which f(a)==0(modq), will also appear at f(-a+1). Thus,
the next occurence of f(x)==0(modq) for positive a<x<a+q appears at
x=q-a+1 (for your example sequence). Note that the two smallest
solutions appear 2a-1 places apart in the sequence (one with negative
index, one with positive index), thus the two smallest solutions with
positive index appear q-2a+1 places apart (which added to a, gives q-
a+1 as stated prior).

Of interest, relative to quadratics with 2nd derivative=2, f(a)*f(a+1)
=f(a+f(a)). Generalizations of this principle to quadratics with 2nd
derivative other than 2, can be made to characterize and calculate
other composites which must appear as members of the quadratic
progression.

> 2,3,5,7,19,29,37,61,71,73,89,113..?? . There are many more, but I
> end it @ prime 113.

These are merely the primes which do not appear in the quadratic
progression as factors. We are guaranteed to have infinitely many
such "missing" prime factors as long as our quadratic progression
does not contain 0 as an integral solution (generally, we also assume
it is not trivially multiplied through by a common factor)
When zero appears in a quadratic progression, all positive integers
appear as factors. To prove this, simply set f(0)=0 where it appears
and then it is always true that f(a)==0(mod ABS[a]).

>
> I am not sure if this is true of false!
>
> If it is true, why?
> Dan

It's all relative and well defined by quadratic reciprocity. In
fact, if you would like to create a quadratic progression for which
11 is the only prime factor less than 113, consider f(x)=x^2-x+11
(your example sequence), and then take three equally spaced solutions
and assign them as three consecutive solutions of a new quadratic
progression given by g(k) such that,
g(-1)=f(1)
g(0)=f(1+m)
g(1)=f(1+2m)

where m = 13*17*23*31*41*43*47*53*59*67*79*83*97*101*103*107*109

Note that the quadratic progression given by g(k) is a subset of that
given by f(x).

The sequence of 3 consecutive seed values completely defines g(k) and
can be expressed in form g(k)=A*k^2+B*k+C by solving
A=(g(-1)-2*g(0)+g(1))/2
B=(g(1)-g(-1))/2
C=g(0)

From there, you could take any g(a) not divisible by 11 and define a
new partial set of the previous progression, and derive a new
quadratic based on consecutive values g(a),g(a+11), g(a+22). This
newly derived quadratic will yield a progression of integers for
which none have a prime factor less than 113.

It's appealing as a means to guarantee a sequence without small
factors, but the values grow enormous quickly, the equations become
unweildly, and in the scheme of things, you're not really improving
your probability of finding a prime by very much.

-Dick
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