A109641 Composite n such that binomial(3n, n) == 3^k (mod n) for some integer k > 0.

4, 9, 15, 25, 27, 34, 36, 49, 51, 57, 63, 68, 75, 81, 87, 93, 111, 121, 125, 129, 132, 138, 141, 153, 155, 159, 169, 177, 237, 249, 258, 261, 264, 267, 274, 276, 279, 289, 298, 303, 324, 339, 343, 357, 361, 375, 381, 387, 393, 411, 417, 423, 441, 447, 453, 477

Offset

1,1

Comments

Includes p^k for k >= 2 and p > 2 in A019334 but not in A014127, as binomial(3n,n) is coprime to p and 3 is a primitive root mod p^k. - Robert Israel, Nov 12 2017

Links

Robert Israel, Table of n, a(n) for n = 1..1000

Example

Binomial(3*34,34) == 3^6 (mod 34), so 34 is a member.

Maple

filter:= proc(n) local p, m, k, t;
  if isprime(n) then return false fi;
  p:= padic:-ordp(n, 3);
  p:= p + numtheory:-order(3, n/3^p);
  m:= binomial(3*n, n) mod n;
  t:= 1;
  for k from 1 to p do
    t:= t*3 mod n;
    if t = m then return true fi;
  od:
false
end proc;
select(filter, [$2..1000]); # Robert Israel, Nov 12 2017

Mathematica

okQ[n_] := Module[{p, m}, If[PrimeQ[n], Return[False]]; p = IntegerExponent[n, 3]; p = p + MultiplicativeOrder[3, n/3^p]; m = Mod[Binomial[3n, n], n]; AnyTrue[Range[p], m == PowerMod[3, #, n]&]];
Select[Range[2, 500], okQ] (* Jean-François Alcover, Mar 27 2019, after Robert Israel *)

Crossrefs

Cf. A019334, A080469, A014127, A109642.

Sequence in context: A348338 A065893 A052116 * A134675 A050530 A278021

Adjacent sequences: A109638 A109639 A109640 * A109642 A109643 A109644

Keyword

nonn

Author

Ryan Propper, Aug 05 2005

Extensions

Corrected and extended by Max Alekseyev, Sep 13 2009

Edited by Max Alekseyev, Sep 20 2009

Status

approved