A212328 Smallest k such that k^3 + 17 is divisible by 3^n.

1, 1, 4, 4, 58, 139, 139, 1597, 1597, 8158, 8158, 67207, 67207, 598648, 2192971, 6975940, 21324847, 21324847, 21324847, 408745336, 1571006803, 8544575605, 29465282011, 29465282011, 217751639665, 500181176146, 1347469785589, 6431201442247, 6431201442247

Offset: 1

Author

Michel Lagneau, May 14 2012

Keywords

• nonn
• easy

Comments

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.

Examples

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

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A129805.

Programs

Maple

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:

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

Extensions

a(20)-a(29) from Charles R Greathouse IV, May 14 2012