Problems & Puzzles: Puzzles

 Puzzle 1049. p=sum of (di^di) Paolo Lava sent the following nice puzzle. Take a prime p and sum its digits each of them powered to itself. If the new number is a prime repeat the process. Here below the list of the least zero-less primes that produce other primes for n steps: n  p 1  11 -> 1^1+1^1 =2. 2  1433 -> 1^1+4^4+3^3+3^3 = 311 -> 3^3+1^1+1^1 = 29. 3  122527 -> 1^1+2^2+2^2+5^5+2^2+7^7 = 826681 -> 8^8+2^2+6^6+6^6+8^8+1^1 = 33647749 -> 3^3+3^3+6^6+4^4+7^7+7^7+4^4+9^9 = 389114797.  Q. Could you find other minimum values for n>3.

Contributions came from Emmanuel Vantieghem, Giorgos Kalogeropoulos and Simon Cavegn.

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Emmanuel, Giorgios and Simon found the same solution for n=4, p=112222268681

112222268681 -> 33647767 -> 2564251 -> 53171 -> 826697 -> 405114564.

But I think it's not minimal (Emmanuel)

It is not difficult (at least in principle) to get a chain of six primes.  Here is one :
12223333445555555555555566666677777778888888888888899999999999999999999989999999
99999999999999999999999999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999 -> 112222268681 -> 33647767 -> 2564251 -> 53171 -> 826697
(-> 405114564  is not prime nor zerofree).

But it is almost impossible to handle the minimal question.

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