Near RepSequence Primes

[davar55]

I tried factoring, with factdb,

123456789*(10^999-1)/(10^9-1)*10+1

i.e. 123456789123456789...1234567891 (1000 digits)

and got an unspectacular P7 * C993.

I tried

123456789*(10^999-1)/(10^9-1)*10+7

and got a P5 * C995.

I don't know if these types of numbers have previously been
labeled, so I made up the thread title to describe prime integers
of the form

k * (10^n*len(k)-1) / (10^len(k)-1) * 10 + d

where d is a decimal digit (0 thru 9).

My puzzle is twofold:

find a near repsequence prime 4-constellation (a+1,a+3,a+7,a+9)
(if that's too hard then a 3- or 2- constellation)
(small or large), and/or

find a big such prime (you know better than I how big is big,
I would say >= 1000 digits).

[science_man_88]

Originally Posted by davar55

k * (10^n*len(k)-1) / (10^len(k)-1) * 10 + d

where d is a decimal digit (0 thru 9).

Originally Posted by davar55

My puzzle is twofold:

find a near repsequence prime 4-constellation (a+1,a+3,a+7,a+9)

this defines a gap [2 4 2] according to your form listed pari finds 5,7,11,13 .

[Batalov]

4690 digits (N-1 is 1% factored; use M.Kamada's Phi pages) <- easy to prove with Primo

14716 digits (N-1 is 6.78% factored)

456789123456...7: 3055 digits 7195 digits 38443 digits (ok, finally something interesting)

234567891...23: 10523 digits

[science_man_88]

Originally Posted by science_man_88

this defines a gap [2 4 2] according to your form listed pari finds 5,7,11,13 .

if n=0 isn't allowed n=1 produces 11,13,17,19 .

[davar55]

Originally Posted by science_man_88

if n=0 isn't allowed n=1 produces 11,13,17,19 .

That constellation does trivially satisfy the formula, but there
are no repeated sequences, so you could try n > 1.

BTW a 2-constellation is a twin prime pair.

[davar55]

Originally Posted by Batalov

4690 digits (N-1 is 1% factored; use M.Kamada's Phi pages) <- easy to prove with Primo

14716 digits (N-1 is 6.78% factored)

456789123456...7: 3055 digits 7195 digits 38443 digits (ok, finally something interesting)

234567891...23: 10523 digits

Nice.

(But I'm not sure the 3055 satisfies the formula, since k=456789123, d=7
should have a length of 9x+1 for the right x. But maybe I'm
misreading your representation.)

[science_man_88]

Originally Posted by davar55

That constellation does trivially satisfy the formula, but there
are no repeated sequences, so you could try n > 1.

BTW a 2-constellation is a twin prime pair.

with n=2 the first group I see is:

1871, 1873, 1877, 1879,

[gd_barnes]

Originally Posted by science_man_88

with n=2 the first group I see is:

SM88, see the puzzle question. It is:

find a near repsequence prime 4-constellation (a+1,a+3,a+7,a+9)
(if that's too hard then a 3- or 2- constellation)
(small or large), and/or

Yours are not near repsequence primes. Quit posting this drivel that has been known for agesl. Good gawd.

[science_man_88]

Originally Posted by gd_barnes

SM88, see the puzzle question. It is:


Yours are not near repsequence primes. Quit posting this drivel that has been known for agesl. Good gawd.

I made a pari script using the form posted:

so I made up the thread title to describe prime integers
of the form

k * (10^n*len(k)-1) / (10^len(k)-1) * 10 + d

using this form:

a=[];for(k=1,999,for(n=2,20,for(d=0,9,if(isprime(k * (10^n*length(k)-1) / (10^length(k)-1) * 10 + d),a=concat(a,k * (10^n*length(k)-1) / (10^length(k)-1) * 10 + d)))));a=vecsort(a,,8)

in the results you will find the four last listed, and the OP has admitted that all ones given before then are of the type they desire.

[Batalov]

I've found some other small rolled-over numbers of which the second largest was 32114 digits, a (234567891)_n23

[davar55]

Originally Posted by Batalov

I've found some other small rolled-over numbers of which the second largest was 32114 digits, a (234567891)_n23

Oh, I see. The light of dawn hits. Please ignore my comment
about your extension of the OP.