A316505 a(n) is the smallest number k > 1 such that k^n - 1 is divisible by 3^n.

4, 8, 10, 80, 244, 242, 2188, 6560, 2188, 59048, 177148, 177146, 1594324, 4782968, 4782970, 43046720, 129140164, 43046720, 1162261468, 3486784400, 3486784402, 31381059608, 94143178828, 94143178826, 847288609444, 2541865828328, 282429536482, 22876792454960, 68630377364884, 68630377364882, 617673396283948

Offset

1,1

Comments

To find such k, let n = 3^t*s, gcd(3,s) = 1. By Euler's totient theorem, k^n - 1 is divisible by 3^n is equivalent to k^s - 1 is divisible by 3^(3^t*s - t). For odd s, the solutions for k are k == 1 (mod 3^(3^t*s - t)); for even s, the solutions are k == +-1 (mod 3^(3^t*s - t)). This gives the formula below in formula section.

Links

Table of n, a(n) for n=1..31.

Formula

a(n) = 3^(n - A007949(n)) - (-1)^n.

Example

For n = 3, k^3 == 1 (mod 27) implies k == 1 (mod 9), so a(3) = 10.
For n = 4, k^4 == 1 (mod 81) implies k == +-1 (mod 81), so a(4) = 80.

Mathematica

Array[3^(# - IntegerExponent[#, 3]) - (-1)^# &, 31] (* Michael De Vlieger, Jul 05 2018 *)

Prog

(PARI) v(n) = valuation(n, 3);
a(n) = 3^(n - v(n)) - (-1)^n;

Crossrefs

Cf. A088032 (conjectured to be the smallest number k > 1 such that k^n - 1 is divisible by 2^n).

Sequence in context: A037004 A136861 A109445 * A293597 A131803 A133270

Adjacent sequences: A316502 A316503 A316504 * A316506 A316507 A316508

Keyword

nonn

Author

Jianing Song, Jul 05 2018

Status

approved