Problems & Puzzles: Puzzles

 Puzzle 652 Recursively Pi+1 = POD(Pi) + 1 POD means here "product of digits". Look at this this example that contains 6 primes: P1=1569889->155521->251->11->2->3 (6 primes total) The sequence of the smallest primes having this property, ordered by the quantity of primes generated is: a(n): 3, 2, 11, 251, 56897, 1569889, ?... for n=1, 2, 3, ... Q. Send your next members for this sequence

Contributions came from W. Edwin Clark, Ryan Bailey, Jan van Delden & Emmanuel Vantieghem

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W. Edwin wrote:

I have verified that for primes p < 10^40 there is NO sequence of 7  primes:

p = P_1 -> P_2 ->...-> P_7

This verification is possible when it is noted that all of the primes
P_2,...,P_7 have to be of the form 2^a*3^b*5^c*7*d + 1. This makes
searches quite efficient since there are so few primes of this form.

I explored the question for other bases a little. My best result was for
base 11 where I obtained the following sequence of 9 primes given in
both their decimal form and in their base 11 form:

P_1 = 953353965375001 = [2, 5, 6, 8, 4, 10, 1, 6, 7, 6, 2, 7, 4, 10, 3]_11
P_2 = 8128512001 = [3, 4, 10, 1, 3, 7, 3, 4, 1, 5]_11
P_3 = 151201 = [10, 3, 6, 6, 6]_11
P_4 = 6481 = [4, 9, 6, 2]_11
P_5 = 433 = [3, 6, 4]_11
P_6 = 73 = [6, 7]_11
P_7 = 43 = [3, 10]_11
P_8 = 31 = [2, 9]_11
P_9 = 19 = [1, 8]_11

My most complete result is that for base 2 there is no sequence with more than 2 primes.
All of the form in binary:   P_1 = 1111...1 -> 10  where P_1 is a Mersenne prime.  :-

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Ryan wrote:

I hope it wasn't just my algorithm, but after checking all primes up to 28204749227, I found no solutions for more than 6 primes.

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Jan wrote:

A number n having digits in the set {1,..,9} generates a POD(n) having prime factors 2,3,5 or 7. The successor of n, i.e. POD(n)+1, will therefore be of the form 2^a*3^b*5^c*7^d+1.
In order for this successor to be prime we should at least have one factor 2 (a>0). If more than one 2 is present we should have at most 1 factor 5 (c<=1) in order to avoid zeroes (the successor will otherwise end with digits 01).

The number 1569889=2^5*3^3*23*79+1 can not have a predecessor since it is not of the given form. In order to find a solution having at least length 7 one should find numbers of the given form generating a list with at least length 6. I tested all potential numbers with 100 digits or less.

We have: (#D=number of digits, #SOL=number of solutions of the required form, ML=maximal length of list found):

#D #SOL  ML
1    3   1
2   13   3
3   30   4
4   31   3
5   47   4
6   53   5
7   56   5
8   56   4
9   54   5
10   61   4
11   56   2
12   68   3
13   74   2
14   77   2
15   63   2
16   57   2
17   70   1
18   39   2
19   52   1
20   44   1
30   26   1
40   13   1
50    7   1
60    0   0
70    4   1
80    0   0
90    0   0
100   0   0
Note that #SOL=3 if #D=1 because 2 is not of the required form. [It could be added if one prefers..].

In fact for #D>=28 I found ML<=1 and even worse ML=0 for #D in {60,62,64,66,67,75,76,77,79,80,81,83,84,85,87,88,90,91,92,93,94,95,96,97,98,99,100}.
So not only do the number of required numbers thin out to 0, the probability that two of these are each other predecessor/successor is close to 0 even worse we need at least 5 connections... So my answer is: there is no next member.

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Emmanuel wrote:

In my opinion, there is no next member in that sequence! I came to that conclusion by the following observations.

Let us abbreviate the function  POD(m) + 1  by  F(m). It is clear that  F(m)  is of the form  1+(2^a) *(3^b)*(5^c)*(7^d).

I examined those  m  for which  F(m)  and  F(F(m))  are prime and for which  F(F(F(m)))  is not 1 (which means that there is no zero among its digits). There were very few candidates among the  m  with less than  200  digits.  All of them lead to chains with at most  6  primes.

So, if there is a next member, it must have more than  200  digits ...

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