- 9839389 = 7 * 43 * 97 * 337

Sum the primes between the smallest

and largest prime factors:

7+11+13+...+331+337 = 10181, a prime.

Sum the composites between the

smallest and largest prime factors:

8+9+10+...+335+336 = 46751, a prime.

Can anyone find larger palindromes

like 9839389?

"In mathematics you don't understand things.

You just get used to them."

--Johann von Neumann - --- In primenumbers@y..., "Max B" <zen_ghost_floating@h...> wrote:

> Sum the primes between the smallest

Yes. They don't seem to be rare.

> and largest prime factors:

> 7+11+13+...+331+337 = 10181, a prime.

>

> Sum the composites between the

> smallest and largest prime factors:

> 8+9+10+...+335+336 = 46751, a prime.

>

> Can anyone find larger palindromes

> like 9839389?

The smallest is 15251=101*151

The sum of the primes

101+103+107+109+113+127+131+137+139+149+151=1367, a prime

the sum of the composites

(151^2-101^2+151+101)/2-1367=5059 a prime

There are more than 300 such palindromes below 2^32

The largest I can conveniently find is

4273223724=2^2*3^2*7*11*467*3301

where the sum of the primes up to and including 3301 is 700897 and

the sum of the composites 4749053 are both primes.

For large solutions to this problem I'd suggest sticking to even

palindromes.

You can divide the problem into two parts. Firstly find pairs of

primes such that the sums of the primes and composites between them

are both primes. Secondly, using the larger prime of the pair and

quite a lot of thought you should be able to solve a degree 1 modular

equation in quite a few variables to generate some candicates you can

then check for smoothness. This could be tricky. One neat way of

generating smooth palindromes is to look for palindromes with low

digit values which are divisible by other, shorter low digit value

palindromes. You can then add them together (with appropriate shifts)

to generate more palindromes of which a large part of the

factorisation is already known. For instance some of the common small

largest factors of palindromes are 101, 9091 (divides 10^5+1), 9901

(divides 10^6+1), 333667 (divides 1001001). Also 3637 but I don't

know why.

If you can find a large factor p of two low weight palindromes of

different sizes such that the sum of the primes and composites from 2

to p are prime then you're probably set for some really big solutions

to this problem.

Richard - Using some of Richard's lucid tips below, and my

usual brute-force blatant bricoleur approach

(too clumsy to divulge here), I managed to find:

819870212078918 =

2 * 7 * 11 * 13 * 17 * 17 * 10771 * 131561

where the sum of primes from 2 to 131561 is

766715749 and the sum of composites is

7887498391, both primes.

But I'm sure this solution could be easily eclipsed

by the 'heavyweights' on this list.

Thanks Richard!

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

From: richard_heylen

To: primenumbers@yahoogroups.com

Sent: Sunday, November 03, 2002 9:02 PM

Subject: [PrimeNumbers] Re: The curious number 9839389.

[...]

The largest I can conveniently find is

4273223724=2^2*3^2*7*11*467*3301

where the sum of the primes up to and including 3301 is 700897 and

the sum of the composites 4749053 are both primes.

For large solutions to this problem I'd suggest sticking to even

palindromes.

You can divide the problem into two parts. Firstly find pairs of

primes such that the sums of the primes and composites between them

are both primes. Secondly, using the larger prime of the pair and

quite a lot of thought you should be able to solve a degree 1 modular

equation in quite a few variables to generate some candicates you can

then check for smoothness. This could be tricky. One neat way of

generating smooth palindromes is to look for palindromes with low

digit values which are divisible by other, shorter low digit value

palindromes. You can then add them together (with appropriate shifts)

to generate more palindromes of which a large part of the

factorisation is already known. For instance some of the common small

largest factors of palindromes are 101, 9091 (divides 10^5+1), 9901

(divides 10^6+1), 333667 (divides 1001001). Also 3637 but I don't

know why.

If you can find a large factor p of two low weight palindromes of

different sizes such that the sum of the primes and composites from 2

to p are prime then you're probably set for some really big solutions

to this problem.

Richard