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

1,1

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

Links

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

Example

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.

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

Prog

(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

Crossrefs

Cf. A161705, A239129.

Sequence in context: A007402 A334505 A375955 * A058260 A202410 A358495

Adjacent sequences: A251728 A251729 A251730 * A251732 A251733 A251734

Keyword

nonn

Author

Michel Lagneau, Dec 07 2014

Status

approved