A251731 Least k such that k^3 + q is divisible by 3^n where q is the n-th number congruent to 1 or -1 (mod 18).

2, 1, 2, 16, 32, 145, 62, 1363, 3458, 19492, 58928, 89308, 70028, 1594318, 1890551, 189871, 31401806, 47918575, 190704887, 163454602, 502048577, 9481323661, 11627845304, 34656488290, 115450061084, 286130228125, 2303721331049, 1569269836240, 22013516320412

Offset: 1

Author

Michel Lagneau, Dec 07 2014

Keywords

• nonn

Comments

It is known that k always exists if q is congruent to +-1 mod 18.

The numbers congruent to 1 or -1 (mod 18) are 1, 17, 19, 35, 37, ... = {A161705} UNION {A239129}.

For n >= 2, k^3 == (9 - 18*n - 7*(-1)^n)/2 (mod 3^n) if and only if k - a(n) is divisible by 3^(n-1). - Jinyuan Wang, Feb 13 2020

Examples

			a(1) = 2 because the first number of the form +-1 (mod 18) is 1, and 2^3 + 1 = 9 = 3*3^1;
a(2) = 1 because the second number of the form +-1 (mod 18) is 17, and 1^3 + 17 = 18 = 2*3^2;
a(3) = 2 because the third number of the form +-1 (mod 18) is 19, and 2^3 + 19 = 27 = 3^3;
a(4)= 16 because the fourth number of the form +-1 (mod 18) is 35, and 16^3 + 35 = 4131 = 51*3^4.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..2095

Crossrefs

Cf. A161705, A239129.

Programs

Maple

f:= proc(n) local q,R,k;
  if n::odd then q:= 9*n-8 else q:= 9*n-1 fi;
  min(map(subs,[msolve(k^3+q,3^n)],k))
end proc:
map(f, [$1..30]); # Robert Israel, Dec 23 2018

Mathematica

lst1={1};Do[lst1=Union[lst1,Union[{18*n+1},{18*n-1}]],{n,1,10}];lst={};Do[k=1;While[Mod[k^3+lst1[[n]],3^n]!=0,k++];Print[n," ",k],{n,1,10}];lst

PARI

a(n) = {if (n % 2, q = 9*(n-1)+1, q = 9*n-1); m = 3^n; k = 1; while ((k^3+q) % m, k++); k;} \\ Michel Marcus, Jan 07 2015