Near RepSequence Primes
 

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).

[QUOTE=davar55;314301]

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

where d is a decimal digit (0 thru 9).[/QUOTE]



[QUOTE=davar55;314301]

My puzzle is twofold:

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

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

[URL="http://factordb.com/index.php?id=1100000000543568421"]4690 digits[/URL] (N-1 is 1% factored; use M.Kamada's Phi pages) <- easy to prove with Primo

[URL="http://factordb.com/index.php?id=1100000000543568487"]14716 digits[/URL] (N-1 is 6.78% factored)

456789123456...7: [URL="http://factordb.com/index.php?id=1100000000543568761"]3055 digits[/URL] [URL="http://factordb.com/index.php?id=1100000000543568960"]7195 digits[/URL] [URL="http://factordb.com/index.php?id=1100000000543574010"]38443 digits[/URL] (ok, finally something interesting)

234567891...23: [URL="http://factordb.com/index.php?id=1100000000543574301"]10523 digits[/URL]

[QUOTE=science_man_88;314308]this defines a gap [2 4 2] according to your form listed pari finds 5,7,11,13 .[/QUOTE]

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

[QUOTE=science_man_88;314386]if n=0 isn't allowed n=1 produces 11,13,17,19 .[/QUOTE]

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.

[QUOTE=Batalov;314330][URL="http://factordb.com/index.php?id=1100000000543568421"]4690 digits[/URL] (N-1 is 1% factored; use M.Kamada's Phi pages) <- easy to prove with Primo

[URL="http://factordb.com/index.php?id=1100000000543568487"]14716 digits[/URL] (N-1 is 6.78% factored)

456789123456...7: [URL="http://factordb.com/index.php?id=1100000000543568761"]3055 digits[/URL] [URL="http://factordb.com/index.php?id=1100000000543568960"]7195 digits[/URL] [URL="http://factordb.com/index.php?id=1100000000543574010"]38443 digits[/URL] (ok, finally something interesting)

234567891...23: [URL="http://factordb.com/index.php?id=1100000000543574301"]10523 digits[/URL][/QUOTE]

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.)

[QUOTE=davar55;314394]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.[/QUOTE]

with n=2 the first group I see is:

[QUOTE]1871, 1873, 1877, 1879,[/QUOTE]

[QUOTE=science_man_88;314408]with n=2 the first group I see is:[/QUOTE]

SM88, see the puzzle question. It is:
[quote]
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
[/quote]

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

[QUOTE=gd_barnes;314423]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.[/QUOTE]

I made a pari script using the form posted:

[QUOTE]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[/QUOTE]

using this form:

[CODE]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)
[/CODE]

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.

I've found some other small rolled-over numbers of which the second largest was [URL="http://factordb.com/index.php?id=1100000000543633098"]32114 digits[/URL], a (234567891)[SUB]n[/SUB]23

[QUOTE=Batalov;314436]I've found some other small rolled-over numbers of which the second largest was [URL="http://factordb.com/index.php?id=1100000000543633098"]32114 digits[/URL], a (234567891)[SUB]n[/SUB]23[/QUOTE]

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