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A182751 a(1)=1, a(2)=3, a(3)=6; a(n) = 3*a(n-2) for n > 3. 8
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

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 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 und 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

LINKS

Harvey P. Dale, Table of n, a(n) for n = 1..1000

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

EXAMPLE

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

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 *)

CROSSREFS

Essentially the same as A038754 (cf. formula).

Cf. A182752 - A182757.

Sequence in context: A007783 A050625 A025614 * A057576 A100852 A059006

Adjacent sequences:  A182748 A182749 A182750 * A182752 A182753 A182754

KEYWORD

nonn,easy,less,changed

AUTHOR

Jaroslav Krizek, Nov 27 2010

STATUS

approved

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Last modified December 5 09:15 EST 2016. Contains 278762 sequences.