A212328 - OEIS

%I #10 May 13 2013 01:54:21

%S 1,1,4,4,58,139,139,1597,1597,8158,8158,67207,67207,598648,2192971,

%T 6975940,21324847,21324847,21324847,408745336,1571006803,8544575605,

%U 29465282011,29465282011,217751639665,500181176146,1347469785589,6431201442247,6431201442247

%N Smallest k such that k^3 + 17 is divisible by 3^n.

%C This sequence is generalizable :  the smallest k such that k^3 + p is divisible by 3^n exists if the prime p is congruent to + - 1 mod 18. For example, the sequence with p = 19 is given by {2, 2, 2, 20, 20, 20, 263, 992, 3179, 16301, 55667, 173765, 528059, …}. (See A129805). This sequence is given with the smallest p = 17.

%H Charles R Greathouse IV, <a href="/A212328/b212328.txt">Table of n, a(n) for n = 1..1000</a>

%e a(4) = 4 because 4^3 + 17 = 81 is divisible by 3^4.

%p with(numtheory):for n from 1 to 20 do:i:=0:for x from 1 to 10^8 while(i=0) do: z:= x^3 + 17:if irem(z,3^n)=0 then i:=1: printf ( "%d %d \n",n,x):else fi:od:od:

%o (PARI) print1(k=1);for(n=2,100,if(Mod(k,3^n)^3!=-17,k+=3^(n-2)* if(Mod(k+3^(n-2),3^n)^3==-17,1,2));print1(", "k)) \\ _Charles R Greathouse IV_, May 14 2012

%Y Cf. A129805.

%K nonn,easy

%O 1,3

%A _Michel Lagneau_, May 14 2012

%E a(20)-a(29) from _Charles R Greathouse IV_, May 14 2012