A091733 - OEIS

%I #17 Apr 25 2016 12:05:00

%S 2,3,4,5,6,7,2,9,4,11,12,13,3,9,16,17,18,7,7,21,4,23,24,25,26,3,10,9,

%T 30,31,5,33,34,35,11,13,10,7,16,41,42,25,6,45,16,47,48,49,18,51,52,9,

%U 54,19,56,9,7,59,60,61,13,5,4,65,16,67,29,69,70,11,72,25,8,47,76,45,23,55,23

%N a(n) is the least m > 1 such that m^3 = 1 (mod n).

%C a(n) <= n + 1; the inequality is strict iff n is divisible by 9 or by a prime congruent to 1 mod 3. - _Robert Israel_, May 27 2014

%H Robert Israel, <a href="/A091733/b091733.txt">Table of n, a(n) for n = 1..10000</a>

%e a(7) = 2 because 2^3 is congruent to 1 (mod 7).

%p A:= n -> min(select(t -> type((t^3-1)/n, integer), [$2 .. n+1]));

%p map(A, [$1 .. 1000]); # _Robert Israel_, May 27 2014

%t f[n_] := Block[{x = 2}, While[Mod[x^3 - 1, n] != 0, x++]; x]; Array[f, 79] (* _Robert G. Wilson v_, Mar 29 2016 *)

%o (MATLAB) m = 2; while mod(m^3 - 1, n); m = m + 1; end; m

%o (PARI) a(n) = my(k = 2); while(Mod(k, n)^3 != 1, k++); k; \\ _Michel Marcus_, Mar 30 2016

%Y Cf. A070667, A076947, A083501.

%K easy,nonn

%O 1,1

%A _David Wasserman_, Mar 05 2004