Problems & Puzzles: Puzzles

Puzzle 112. Automorphic primes

I.- The d->(d)d transformation

A few days ago G. L. Honaker, Jr. was re-exploring the Conway's Look and say - sequence and had the idea of create sequences of primes by the simple procedure of replacing digits using certain specific transformations.

At the beginning he selected the following procedure: replace the digit 'd' with d digits 'd'. Do this for all the digits of the current number.

Then he started asking & getting primes (non trivial) that generates other primes?

If we restrict to only one iteration of the procedure the answer is 31:

31 --> 3331

He also found the least prime that provides a sequence of 3 primes, applying twice the procedure:

641--> 66666644441--> (6)36(4)161

I (C.R.) found the least prime (12422153) that starts a sequence of 4 primes and Paul Jobling has produced a prime (142112242123, not necessarily the least) that starts a sequence of 5 primes

Then we have the following sequence:

2, 31, 641, 12422153, 142112242123 , ...

Whose description is:

"The least prime that starts a sequence of k primes using recursively the d->(d)d transformation k-1 times".

Questions:

1. Can you confirm if the Jobling's prime (142112242123) is the 5th member of the d->(d)d sequence, or to get the least if this is not the case?
2. Can you get the 6th member of the sequence
d->(d)d ?

II.- The d->d2 transformation

More recently G. L. Honaker asked for the corresponding sequences for the following transformation d->d2. Himself sent the 131 -> 191 -> 1811 -> 16411 (4 primes) example. I produced the following two examples:

a) 2111-> 4111-> 16111-> 136111-> 1936111 (5 primes) and

b) 12815137 -> 14641251949 -> 116361614251811681 -> 11369361361164251641136641 -> 11936819361936113616425136161193636161 ->
1181936641819361819361193613616425193613611819369361361 (6 primes)

Then, this is the corresponding sequence:

2, 13, 13, 131, 2111, 12815137, ?

Question 3: Can you extend the d->d2 sequence?

III.- The d->dd transformation

Very naturally I extended the idea of G. L. Honaker making the power of the digit d to be the same digit d, that is to say, d->dd

This is the corresponding sequence:

2, 13, 367, 8071171, ?, ...

In this transformation:
13 -> 127

367-> 2746656823543 -> 4823543256466564665631254665616777216427312525627

and

8071171 -> 167772161823543118235431 ->
146656823543823543823543414665611677721642731252562711167772164273125256271 ->
125646656466563125466561677721642731252562716777216427312525627167772164273/
12525627256125646/656466563125466561146656823543823543823543414665625648235/
43271431254312546656482354311146656/823543823543823543414665625648235432714/
3125431254665648235431

Question 4: Can you extend the d->dd sequence?

As per Question 2 Felice Russo wrote (20/11/2000): "After 1 week of search I was not able to find any further solution between 12815137 and 63*10^8."

***

Eleven years later (Set, 2011) Giovanni Resta found that the Jobling's result for Q1 was wrong and more issues:

Part I (Question 1). The 5th member in the
sequence is not Jobling's 142112242123.
Up to it, the initial values that start a sequence of 5 primes  are
the following (several are obtained from the first by inserting zeros):

Number ---- Digits of last prime
12505785661 [10982]
12550785661 [10982]
12557085661 [10982]
12557856601 [10982]
66132153133 [3560]
102550785661 [10982]
102557085661 [10982]
125050785661 [10982]
125507856601 [10982]
125578500661 [10982]
132471145421 [3911]
142112242123 [677] (Jobling's)

So, if zero is permitted the 5th member of the sequence
is 12505785661, if not, it is 66132153133.

Part II (Question 3). The 6th member of the sequence
is 103723971119 which produces a last prime of 102 digits.
The prime produced are:
1094949814911181,
1081168116816411681111641,
106411366411366413616113664111136161,
103616119363616119363616193613611936361611119361361,
10936136118193693613611819369361361819361936118193693613611118193619361,
1081936193611641819368193619361164181936819361936164181936181936116418
19368193619361111641819361819361.

Part III (Question 4). The 5th member in the
sequence is  137101333, which leads to a last
prime of length 525. (Here I assumed that 0^0=1, as
it has been done for the 4th member of the sequence).

I have seen the large of the primes for your answers in Q1. Are the last primes rigorous prime numbers (CR)?

No, they are just probable primes, according to Mathematica and according to
the primality test  in the multiple precision library GMP and also according to the program Pfgw. I think that obtaining a certificate for a prime of 10000 digits would require a very long time on my machine. Anyway, I'm not a fan of large primes, so I'm not an expert on the software needed to certify them.

***

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