[ July 7, 2002 ]
Primes remaining prime when its digits are intertwined
with their SOD (Sum Of Digits) strings

For all I know 30103 is the smallest such palprime
with a palindromic SOD as well ie. 7 so after
intertwining I get this new palprime 370717073.

Alas, both 19 or 30103 are not good examples
when I want to introduce iteration to the operation.
19 yields 1109 but 1109 yields 1111110119
which is composite (7 * 9283 * 17099) unfortunately.
30103 yields 370717073 but 370717073 yields
3357350357351357350357353
which is neither prime nor palindromic as this number
equals 3^3 * 7 * 79 * 3648989 * 61621918451567.

It shouldn't be hard to figure out the next question now !
Find the smallest primes (and/or palprimes) that remain prime
(and/or palprime) when applying the described procedure
after two or more iterations.

Note : the table will be filled with the first primes
to have ONLY 'i' iterations.

iterations	prime palprime important note from JCR	who
1	19 30103	pdg
2	5003 ? (at least 23 digits)	jh jcr
3	3023	jh
4	? (larger than 10,000,000)	jh
5	?	?
6	?	?

Jeff Heleen found the solution for iteration 3
(a total of 4 primes) [ July 14, 2002 ]

3023
3808283
3328320328322328323
3623622628623622620623622628623622622623622628623622623
alas
320662062206320662062... ...620622062206620622063
is composite !

On [ July 15, 2002 ] Jeff wrote
" I have taken my search up to 10^5. I used a modified ECM program
for the search, which was partly manual. Where the numbers got too
large for the ECM program to handle I then used Primo.exe to finish.
I found a total of 3 numbers that went up to iteration 3. The lowest,
as mentioned, is 3023. There were 96 numbers that went up to iteration 2.
And there were many many more (which I did not record) that only went
to the first iteration after the initial number.
If/when I refine the program to where it is totally automated I may
continue into higher ranges."

Iteration 3 sequence 3023, 12799, 18353, ...

On [ July 16, 2002 ] Jeff wrote
" I have found that 5003 is the first prime to have ONLY two iterations.

5003 = prime
5808083 = prime
5328320328320328323 = prime

This should fill in another blank in your table.
Also, I have continued the search in larger numbers. From 100000 to
999999 there are no solutions of any sort as all iterations after the
first prime are divisible by 3.

In my search from 1E6 to 2E6 I have only found one instance where
there are 3 iterations (ie, 4 primes).
The starting prime for that is 1915411.
The program (ECM) is mostly automated now but testing the third
iteration (fourth number) must be done manually in Primo. So it is
quite time consuming. I'll continue further."

On [ July 23, 2002 ] Jeff Heleen has an update

" I have searched up to 10,000,000 (1E7) and have
not found any start primes that yield greater than 3 iterations."

[ July 19-22, 2002 ]
Jean Claude Rosa (email) sent 'important' information
regarding the palprimes of this WONplate.

" There exist no palprimes of 3, 5, 7, 9, 11 and 13 digits
that could be the startprime for a second iteration.
I am sorry for this negative result."

Here are some arguments for the proof.

a) The prime numbers of 3, 6, 9, 12, 15, ... digits don't satisfy
because after intertwining the SOD (Sum Of Digits)
one gets a number always divisible by 3. So what is left
are the primes of 5, 7, 11, 13, 17, ... digits.

b) Let PP be the starting palprime;
Let NPP be the palprime obtained after one iteration and
let NNPP be the palprime obtained after the second iteration.
It is self-evident that for NPP to be palindromic
so must SOD(PP) be palindromic
and the same applies for NNPP and SOD(NPP).

Let us illustrate this scheme of the proof with an example:

Take a PP with 7 digits or PP = abcdcba
For SOD(PP), 2 cases are possible
(from 13 digits on, 3 cases are distinguishable).

1) SOD(PP) ≥ 2 and SOD(PP) < 10.
Let SOD(PP) = s (s is a digit), thus s < 10 and s = a+b+c+d+c+b+a
Next we get NPP = asbscsdscsbsa, so SOD(NPP) = 7 * s.
But for NNPP to be palindromic so must SOD(NPP) also be palindromic
leaving us with one solution namely SOD(NPP) = 77, whence s = 11
which contradicts with our starting departure that s < 10 !

2) SOD(PP) ≥ 10 and SOD(PP) = xx (x means a digit).
NPP = axxbxxcxxdxxcxxbxxa whence SOD(NPP) = 23 * x.
There is only one solution possible SOD(NPP) = 161
with s = 7 therefore SOD(PP) = 77
but this is impossible since the maximum of SOD(PP) = 63.

The same method demonstates that for a PP with 11 digits
there are no solutions.

Let us now look at the case with 13 digits.

1) SOD(PP) < 10, same result as above
2) SOD(PP) = xx, whence SOD(NPP) = 35 * x, no palindromic result
3) SOD(PP) = xyx, 2 solutions :
SOD(PP) = 101 and SOD(PP) = 111 (maximum sum is 117 = 13 * 9).

SOD(NPP) = 125 * x + 22 * y
if x = 1 and y = 0 then SOD(NPP) = 125
but this is not a palindrome
if x = 1 and y = 1 then SOD(NPP) = 147
which is neither a palindrome !

To conclude we can state that there is
no second iteration for palprimes with 13 digits.

Furthermore J. C. Rosa wrote:
Two messages : one is  bad , the other is  good  !

" There exist no palprimes of 17 or 19 digits
that could be the startprime for a second iteration."

" There may exist a palprime solution starting with
23 digits if the SOD equals from 7 up to 11.
All that is left now is to discover this palprime !"