>

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

> Primes of this form --

>

> 11 +2,+4,+6,+8,+10,+12,+14,..+n

>

For any solution f(a)=p*q, there will exist the solution sequences f

> Where if at any point the sum is a composite then these composites

> will not have any of these primes in their prime factors.

(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

second derivative of the quadratic.

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

These are merely the primes which do not appear in the quadratic

> end it @ prime 113.

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]).

>

It's all relative and well defined by quadratic reciprocity. In

> I am not sure if this is true of false!

>

> If it is true, why?

> Dan

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