|
| |
|
|
A087094
|
|
a(n) = smallest k such that (10^k-1)/9 == 0 mod prime(n)^2, or 0 if no such k exists.
|
|
4
|
|
|
|
0, 9, 0, 42, 22, 78, 272, 342, 506, 812, 465, 111, 205, 903, 2162, 689, 3422, 3660, 2211, 2485, 584, 1027, 3403, 3916, 9312, 404, 3502, 5671, 11772, 12656, 5334, 17030, 1096, 6394, 22052, 11325, 12246, 13203, 27722, 7439, 31862, 32580, 18145, 37056, 19306
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
For a given a(n)>0, all of the values of k such that (10^k-1)/9=0 mod prime(n)^2 is given by the sequence a(n)*A000027, i.e. integral multiples of a(n). For example, for n=2, prime(2)=3, a(n)=9, the set of values of k for which (10^k-1)/9=0 mod 3^2 is 9*A000027=9,18,27,36,45,...
The union of the collection of sequences formed from the nonzero terms of a(n)*A000027, gives the values of k for which (10^k-1)/9 is not squarefree, see A046412. All of terms of the sequence a(n) are integer multiples of prime(n) for primes <1000 except for a(93)=486 where prime(93)=487. Conjecture: there are no 0 terms after a(3).
That conjecture is easily proved, for a(n) is just the multiplicative order of 10 modulo (prime(n))^2 for n>3. - Jeppe Stig Nielsen, Dec 28 2015
|
|
|
LINKS
|
|
|
|
FORMULA
|
|
|
|
EXAMPLE
|
a(2)=9 since 9 is least value of k for which (10^k-1)/9=0 mod 3^2.
|
|
|
MAPLE
|
0, 9, 0, seq(numtheory:-order(10, ithprime(i)^2), i=4..100); # Robert Israel, Dec 30 2015
|
|
|
PROG
|
(PARI) a(n)=p=prime(n); 10%p==0 && return(0); for(k=1, p^2, ((10^k-1)/9) % p^2 == 0 && return(k)); error() \\ Jeppe Stig Nielsen, Dec 28 2015
(PARI) a(n)=p=prime(n); if(10%p==0, 0, 10%p==1, 9, znorder(Mod(10, p^2))) \\ Jeppe Stig Nielsen, Dec 28 2015
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
