|
%I #14 Dec 31 2015 02:15:17
%S 0,9,0,42,22,78,272,342,506,812,465,111,205,903,2162,689,3422,3660,
%T 2211,2485,584,1027,3403,3916,9312,404,3502,5671,11772,12656,5334,
%U 17030,1096,6394,22052,11325,12246,13203,27722,7439,31862,32580,18145,37056,19306
%N a(n) = smallest k such that (10^k-1)/9 == 0 mod prime(n)^2, or 0 if no such k exists.
%C 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,...
%C 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).
%C 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
%H Robert Israel, <a href="/A087094/b087094.txt">Table of n, a(n) for n = 1..10000</a>
%F For n>3, a(n) = A084680(prime(n)^2) = A084680(A001248(n)), _Jeppe Stig Nielsen_, Dec 28 2015
%e a(2)=9 since 9 is least value of k for which (10^k-1)/9=0 mod 3^2.
%p 0,9,0,seq(numtheory:-order(10,ithprime(i)^2), i=4..100); # _Robert Israel_, Dec 30 2015
%o (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
%o (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
%Y Cf. A000040, A000042, A046412, A084006, A084007.
%K nonn
%O 1,2
%A _Ray Chandler_, Aug 10 2003
|
