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A182751
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a(1)=1, a(2)=3, a(3)=6; a(n) = 3*a(n-2) for n > 3.
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11
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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
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OFFSET
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1,2
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COMMENTS
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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 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
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LINKS
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FORMULA
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a(2*k) = (3/2)*a(2*k-1) for k >= 2, a(2*k+1) = 2*a(2*k).
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EXAMPLE
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For n = 5; a(3) = 6, a(4) = 9, a(5) = 18 before ((6+9)*(6+18)*(9+18)) / (6*9*18) = 10.
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MATHEMATICA
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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 *)
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PROG
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(PARI) x='x+O('x^30); Vec(x*(1+3*x+3*x^2)/(1-3*x^2)) \\ G. C. Greubel, Jan 11 2018
(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
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CROSSREFS
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Essentially the same as A038754 (cf. formula).
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KEYWORD
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nonn,easy,less
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AUTHOR
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STATUS
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approved
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