Problems & Puzzles: Puzzles

Puzzle 851. Puzzle 847 revisited In the Puzzle 847 Abhiram R. Devesh asked for large runs of consecutive prime numbers with the same length of its respective Collatz sequences. Successive puzzlers got new larger runs. Who Run of intergers with the same length Starting prime Last prime Common length Abhiram R. Devesh 22 27512549 27512879 224 Vicente F. Izquierdo 33 79328611 79329179 149 Emmanuel Vantieghem 41 734483273 734484169 190 Michael Hürter 269 3416702480489 3416702488091 334 Dmitry Kamenetsky 15733190 integers (with 36386 primes inside) 2^626+1 (2^262+483) 2^626 + 15733190 (2^626+15733179) 5212 Dmitry found inspiration for his search for large runs in the paper by  Guo-Gang Gao, On consecutive numbers of the same height in the Collatz problem, as Dmitry reported in his own sequence A277109. Well done by Dmitry. But now we have new questions. Here are his new questions on the same original issue of Puzzle 847: 1. What is the longest run you can obtain using the pattern 2^n+1? 2. Why does the integers 2^n+1 give such long runs starting point, as 2^626+1 and many others integers with the same structure do? 3. Are there any other structure of integers that give long runs starting points?

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Contributions came from Emmanuel Vantieghem

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Emmanuel wrote:

Q1.  In my opinion, Dmitry's record will stand for a very long time !

 

Q2.  I think long runs are not especially obtained by numbers of the form  2^n + 1.

Let us denote the maximal number of consecutive integers with the same Collatz length as n  by  L(n).

I found  1158  values of  n  < 2000  for which   L(2^n + 1) < 31.

For  3 <  n < 22, the maximum of  L(m)  when  2^n < m <2^(n+1)  is never obtained when  m = 2^n + 1.  But this can be due to the fact that this deals with small numbers.

 

Q3.  I tried numbers of the form  3*2^n + 1  for  n < 468.  I found  L( 3*2^n + 1) = L(2^n + 1)  with only three exceptions : n = 17, 67  and 68.

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Dmitry Kamenetsy wrote on Oct 26, 2016

I have a new record. I found that every number from 2^1812+1 to 2^1812+2^26-1 has a Collatz height of 12969. In that range I found about 53229 primes from 2^1812+115 to 2^1812+67105183. Note that I can't check them for sure due to their large size.

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On October 30, Carlos Rivera wrote:

I have computed the quantity of strong-pseudoprimes in the range 2^1812+1 to 2^1812+2^26-1. My result is equal to the quantity given by Dmitry: 53229

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