Problems & Puzzles: Puzzles

Puzzle 779. A first follow-up to Puzzle 776

This follow up is a result of the search of Jan van Delden who persisted seeking for more than 10 consecutive integers satisfying the conditions of the puzzle 776.

He has not only succeeded in this target but he also has shown that the first integer of the sequence is not obliged to be 0 mod 10, as the original results may suggest.

Here is the whole follow up as written by Jan:

"In Puzzle 776 we asked for a starting integer N with a transformation M(N)=N+sum(d[i]^2) with d[i] equal to the digits of N, such that M(N[j]) is prime for N[j]=N+j, where j in [0..length-1].

As Giovanni Resta showed in his answer to Puzzle 776, there are a lot of solutions where the last digit of N equals 0 ("711 such numbers up to 2*10^14").

However there also exist solutions with different ending digits.

 
A (noncomplete) check shows:
 

N

length

10

10

172640493891

10

303982397892

10

1582259366593

10

128141028594

9

3050246869595

10

51636003896

9

974882781297

9

8926958015598

9

318532696599

8

 
Solutions with length>10 are also possible:
 

N

length

178209124090

(Giovanni Resta)11

5462349274891

11

4057729195597

11

 
Q1: Find solutions for each last digit, having greater length.
Q2: What is the maximum length, given the last digit?"


Jan van Delden contributed new-better results on March 27, 2015

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Q1: My smallest solution N (N<=5.10^14, given maximum length found) and ending patterns (*):

N

Length

178209124090

11

5462349274891

11

82589451476592

11

72033742697893

11

9189680586294

10

144191157980895

12

19476180832596

10

4057729195597

11

489644422253598

12

172443421786599

11

 

Q2: As written in puzzle 776 a starting value N induces a list of differences between M[i]-M[1]. Such a list is an admissible constellation if not all residues mod p are present, with p<=length(M). If one computes the maximal length L, i.e. for  L+1 there are no free residue anymore. For ending digit patterns 000 through 999 one finds that:

Ending digit + L <= 20

So the given solution ending with digit 8 is maximal.

(*) Digit patterns which give solutions with a length > 10:

19x,49x,79x with x equal to 8 or 9.
29x,59x,89x  with x any digit.

And:

090,390,690

I didnít check these last 3 patterns because Giovanni checked these already and moreover for these patterns the maximal length equals 11.

Note: for 29x,59x,89x I forgot to check a few residues, so this table might not be final, however the results with ending digits 0..3 are minimal.

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