A191627 a(n) = floor(3^n/(3n-1)).

1, 1, 3, 7, 17, 42, 109, 285, 757, 2036, 5535, 15184, 41955, 116657, 326111, 915887, 2582803, 7309820, 20754669, 59098040, 168715374, 482785532, 1384458512, 3977880795, 11449846073, 33011244523, 95319968562, 275624005481, 798027643777, 2313383506681

Offset

1,3

Comments

Also equal to floor((3^n+1)/(3n-1)). [Proof: 3^n (mod 3n-1) is never zero; therefore 3^n -(3^n (mod 3n-1)) equals (1+3^n)-((1+3^n) mod (3n-1)), and removing the remainders in the numerator of either 1+3^n or 3^n results in the same number. The proof that 3^n <> 0 (mod 3n-1) follows from assuming that 3^n=k*(3n-1) for some integer k; reading this requirement mod 3 induces k=3*k', so 3^n=3*k'*(3n-1); reading this mod 9 induces k'=3*k'' etc. ad infinitum through all powers of 3, and forcing k->infinity.] - R. J. Mathar, Jun 20 2011

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..600

Mathematica

f[n_] := Floor[(3^n)/(3 n - 1)]
Table[f[n], {n, 1, 40}]   (* A191627 *)

Prog

(PARI) a(n)=3^n\(3*n-1) \\ Charles R Greathouse IV, Jun 17 2011
(Maxima) A191627(n):=floor( 3^n/(3*n-1) )$ makelist(A191627(n), n, 1, 30); /* Martin Ettl, Oct 25 2012 */

Crossrefs

Cf. A191628, A191629, A191631.

Sequence in context: A244455 A102071 A363142 * A178778 A335596 A238824

Adjacent sequences: A191624 A191625 A191626 * A191628 A191629 A191630

Keyword

nonn,easy

Author

Clark Kimberling, Jun 09 2011

Status

approved