%C A136296 "Special augmented numbers" from Zak Seidov

[Alexander Povolotsky]

What about relation between A136296 and A116436 ?


A116436  Numbers n which when sandwiched between two 1's give a multiple of n.
1, 11, 13, 77, 91, 137, 9091, 909091, 5882353, 10989011, 12987013,
52631579, 76923077, 90909091, 4347826087, 9090909091, 13698630137,
909090909091, 3448275862069, 10989010989011, 12987012987013,
76923076923077 (list; graph; listen)
OFFSET

1,2
EXAMPLE

77 belongs since 1771 is a multiple of 77 (77*23).
MATHEMATICA

f[k_, d_] := Flatten at Table[Select[Divisors[k*(10^(i + 1) + 1)],
IntegerLength[ # ] == i &], {i, d}]; f[1, 14] (*Chandler*)
CROSSREFS

Cf. A116437-A116444.

Sequence in context: A057189 A105290 A072580 this_sequence A056446
A023317 A056436

Adjacent sequences: A116433 A116434 A116435 this_sequence A116437
A116438 A116439
KEYWORD

base,nonn
AUTHOR

Giovanni Resta (g.resta(AT)iit.cnr.it), Feb 15 2006

pa

On 4/22/08, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:
> Definitely, this morning I'm a bit dizzy...
> I agree that your code is much more intelligent.
> (but the existing sequences only lists primes, so "divisors" should be
> replaced by "factor()[,1]" ;
> also I don't think there can be more than the largest and second
> largest in the prime factors to have the required number of digits)
> Maximilian
>
>
> On Tue, Apr 22, 2008 at 11:06 AM,  <franktaw at netscape.net> wrote:
> > This is not so difficult.  We're looking for k digit numbers that divide
> >  10^{k+1}+1.  (Anybody not see why this is equivalent?)
> >
> >  Some PARI:
> >
> >  A136296k(k) = {local(l, d, lb, ub);
> >  d=divisors(10^(k+1)+1);l=[];
> >  lb=10^(k-1);
> >  ub=10*lb;
> >  for(i=1,#d,if(d[i]>=lb&&d[i]<ub,l=concat(l,[d[i]])));
> >  l}
> >
> >  l=[];for(i=1,20,l=concat(l,A136296k(i)));l
> >
> >  And we get:
> >
> >  1, 11, 13, 77, 91, 137, 9091, 909091, 5882353, 10989011,
> >  12987013, 52631579, 76923077, 90909091, 4347826087,
> >  9090909091, 13698630137, 909090909091, 3448275862069,
> >  10989010989011, 12987012987013, 76923076923077,
> >  90909090909091, 9090909090909091, 909090909090909091,
> >  1369863013698630137, 10989010989010989011,
> >  12987012987012987013, 20408163265306122449,
> >  76923076923076923077, 90909090909090909091
> >
> >  Franklin T. Adams-Watters
> >
> >
> >
> >  -----Original Message-----
> >  From: Maximilian Hasler <maximilian.hasler at gmail.com>
> >
> >  1p1 divisible by p
> >  <=> 10^[log10(p)+2] + 10p + 1 divisible by p
> >  <=> 10^[log10(p)+2] + 1 divisible by p
> >  <=> 10^[log10(p)+2] = -1 (mod p)
> >
> >  Three more terms ; a(7)>10^7.
> >
> >  %I A136296
> >  %S A136296 11, 13, 137, 9091, 909091, 5882353
> >  %F A136296 1p1 divisible by p <=> 10^[log[10](p)+2] = -1 (mod p)
> >  %o A136296 (PARI) forprime(p=1,10^7,Mod(10,p)^(log(p)\log(10)+2)+1 |
> >  print1(p", "))
> >  %A A136296 M. F. Hasler (Maximilian.Hasler at gmail.com), Apr 22 2008
> >
> >
> >  On Tue, Apr 22, 2008 at 1:24 AM, zak seidov <zakseidov at yahoo.com> wrote:
> >
> > > %C A136296 "Special augmented numbers"  p such that
> > >  the decimal number 1p1 is divisible by p:
> > >
> > >
> >  1,11,13,77,91,137,9091,909091,90909091,9090909091,909090909091,9090909090
> >  9091,9090909090909091,909090909090909091.
> >
> > >  Notice the infinite pattern
> > >  p=(90..90..90)91 with 1p1/p=21, e.g.,
> > >  1911/91=190911/9091=19090911/909091=21.
> > >
> > >  Prime numbers are
> > >  11, 13, 137, 9091, 909091, 909090909090909091
> > >
> >
> >
> >
> >
> >
> >
>