Problems & Puzzles: Puzzles

Puzzle 1023. Another ladder of primes...

JM Bergot sent the following nice puzzle:

Take prime 53 to see that the sum of its
digits is 8, so 53 base 8 is 43; 43 base 7
is 31; 31 base 4 is 13; and 13 base 4 is
7 STOP at single digit prime. 

Q. Can you find a longer list using a different starting prime?

From Nov 28 to Dec 4, contributions came from  Giorgos Kalogeropoulos, Oscar Volpatti, Metin Sariyar


Giorgios wrote:

In each step, if the starting number is N=a(1)a(2)...a(n) and the next is N' then
N'=a(1)*(a(1)+a(2)+...+a(n))^(n-1) + a(2) *(a(1)+a(2)+...+a(n))^(n-2) + ... + a(n)
So, in order N'<N the sum of digits of N must be less than 10: a(1)+a(2)+...+a(n)<10
  I believe that the longest descending lists containing only primes are
  {53,43,31,13,7} and {61,43,31,13,7}    
We can find bigger lists starting with a prime if not all the terms are be prime:
{223,115,61,43,31,13,7} has more steps but 115 is not a prime
The biggest starting prime that produced a list is:
{11200001,100001,33,21,7} again not all terms are primes
All primes less than 10^20 with sum of digits less than 10 were checked with no results...


Oscar wrote:

If I properly understood the recursion step, we can get an infinite sequence by choosing a starting prime with digit sum 10:
19, 19, 19, 19...
37, 37, 37, 37...  
and so on.
I excluded such "easy" solutions by adding a simple constraint:
within a legal list, consecutive elements must be different.
This is the first list of primes with length 6:
2693, 18583, 518953, 144316411, 178008299401, 4481394332797755409.
For starting primes p < 5.8*10^11, there are about sixty lists with length 6 and no longer ones. 

Let's consider a starting prime with digit sum 100:
699998998879, 60909090908090908080709, 600090009000900090008000900090008000800070009...
and so on; at each step, we find the same non-zero digits, but the spacing between them is doubled; in this case, only the first three terms are prime.

Five starting primes p < 3.8*10^14 generate lists with length 6:
Again, no longer lists found so far.


Metin wrote:

The result I found has 8 steps :
2000130173 =>55521311=>949651=>24691=>2707=>1093=>661=>421

BTW, Metin wrote that the example given by Bergot is incorrect, because 53 base 8 is not 43.


Here are contributions from Metin & Oscar, sent during the week 4-11 December, 2020, explaining the distinct results sent to this puzzle:

Metin wrote:

The question is not open enough I think. My result is true for another understanding, that's all.
I see that Bergot means 53 from sum of its digits to base 10 ? because for only , 53_8 in base 10 =43 .
What I understood was take 53 as a number in base 10, and convert it to base 8 because its sum of digits in base 10 is 8 . So 53 in base 8=65. I found my results in this way, and with every step I convert the new obtained number as base 10 and converted  to its sum of digits in base 10. Everybody may be right by their understanding the problem. 


Oscar wrote:

I rewrote my explanation about puzzle 1023, I hope it's simpler and clearer now.
I also double-checked Giorgos' search for strictly decreasing solutions to Bergot's puzzle.
I didn't work on Metin's puzzle so far, just in case you'll decide to use it as a follow-up puzzle.


JM Bergot and Metin Sariyar understood the words "53 base 8" in different ways, so they actually worked on different puzzles.

Metin's puzzle.
x has base-10 expansion A = [a(1),...,a(m)], with digit sum s = a(1)+...+a(m);
x also has base-s expansion B = [b(1),...,b(n)];
let x' be the unique integer with base-10 expansion B.
If B is not a legal base-10 expansion, fix it by carrying base-10;
it may happen when s > 10 and some b(k) are greater than 9.

x = 55521311
A = [5,5,5,2,1,3,1,1]
s = 5+5+5+2+1+3+1+1 = 23
B = [8,14,9,6,5,1]
B' = [9,4,9,6,5,1]
x' = 949651

Bergot's recursion.
again, x has base-10 expansion A = [a(1),...,a(m)], with digit sum s = a(1)+...+a(m);
let x' be the unique integer with base-s expansion A.
If A is not a legal base-s expansion, then stop;
it happens when s < 10 and x = s*10^(m-1), so that the only non-zero digit in A is a(1) = s.

x = 53
A = [5,3]
s = 5+3 = 8
x' = 5*8+3 = 43

If we allow to fix A by carrying base-s, we get x' = s^m. But:
for m > 1, both x and x' are composite;
for m = 1, equality x' = s = x holds and we reach a repeating sequence;
so, when x is a single digit prime, it's better to stop.  

Both recursions are ill-defined for numbers x = 10^k, as s = 1 and we should work with base-1 representations.
Both recursions generate a repeating sequence x,x,x,... for numbers x with digit sum s = 10.


Giorgos' puzzle.
Bergot's example sequence has two nice properties:
1) it contains only primes;
2) it's strictly decreasing.

I focused on property 1 only, and I found some strictly increasing sequences with length 6.
On the other hand, Giorgos focused on property 2, eventually relaxing primality constraint.

This version of Bergot's puzzle seems much harder, because large numbers are very likely to have digit sum greater than 10, so that x' > x.
We can easily find sequences of two large primes, like:
But what about longer sequences? 
Working backwards, I checked all sequences ending with numbers below 2.8*10^32.
Even after completely dropping primality and allowing carrying base-s, I only found 409 sequences with length at least 3, and none with length greater than 8.

Only five known sequences contain five primes or more (composites in red):

n,p:  sequence
6,5:  203,53,43,31,13,7.
7,5:  215,141,61,43,31,13,7.
7,6:  223,115,61,43,31,13,7.
7,6:  1031,141,61,43,31,13,7.
8,5:  13010,1005,221,61,43,31,13,7.

Largest known prime starting a sequence with length at least 3:

Largest known composite starting a sequence with length at least 3:


Records   |  Conjectures  |  Problems  |  Puzzles