A182751 a(1)=1, a(2)=3, a(3)=6; a(n) = 3*a(n-2) for n > 3.

1, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, 729, 1458, 2187, 4374, 6561, 13122, 19683, 39366, 59049, 118098, 177147, 354294, 531441, 1062882, 1594323, 3188646, 4782969, 9565938, 14348907, 28697814, 43046721, 86093442, 129140163, 258280326, 387420489

Offset: 1

Author

Jaroslav Krizek, Nov 27 2010

Keywords

• nonn
• easy
• less

Comments

For n >= 3: a(n) = the smallest number > a(n-1) such that ((a(n-2) + a(n-1))*(a(n-2) + a(n))*(a(n-1) + a(n)))/(a(n-2)*a(n-1)*a(n)) is an integer (= 10 for n >= 4).

Number of necklaces with n-1 beads and 3 colors that are the same when turned over and hence have reflection symmetry. Example: For n=4 there are 9 necklaces with the colors A, B and C: AAA, AAB, AAC, ABB, ACC, BBB, BBC, BCC, CCC. The only necklaces without reflection symmetry are ABC and ACB. - Herbert Kociemba, Nov 24 2016

Examples

			For n = 5; a(3) = 6, a(4) = 9, a(5) = 18 before ((6+9)*(6+18)*(9+18)) / (6*9*18) = 10.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (0,3).

Crossrefs

Essentially the same as A038754 (cf. formula).

Cf. A182752 - A182757.

Programs

Magma

I:=[3,6]; [1] cat [n le 2 select I[n] else 3*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 11 2018

Mathematica

Join[{1},RecurrenceTable[{a[2]==3,a[3]==6,a[n]==3a[n-2]},a[n],{n,50}]] (* or *) Transpose[NestList[{#[[2]],#[[3]],3#[[2]]}&,{1,3,6},49]][[1]] (* Harvey P. Dale, Oct 19 2011 *)
Rest@ CoefficientList[Series[x (1 + 3 x + 3 x^2)/(1 - 3 x^2), {x, 0, 34}], x] (* Michael De Vlieger, Nov 24 2016 *)
Join[{1}, LinearRecurrence[{0, 3}, {3, 6}, 30]] (* Vincenzo Librandi, Nov 25 2016 *)

PARI

x='x+O('x^30); Vec(x*(1+3*x+3*x^2)/(1-3*x^2)) \\ G. C. Greubel, Jan 11 2018

Formula

a(n) = A038754(n) for n >= 2.

a(2*k) = (3/2)*a(2*k-1) for k >= 2, a(2*k+1) = 2*a(2*k).

G.f.: x*(1 + 3*x + 3*x^2)/(1 - 3*x^2). - Herbert Kociemba, Nov 24 2016