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A027472
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Third convolution of the powers of 3 (A000244).
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43
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1, 9, 54, 270, 1215, 5103, 20412, 78732, 295245, 1082565, 3897234, 13817466, 48361131, 167403915, 573956280, 1951451352, 6586148313, 22082967873, 73609892910, 244074908070, 805447196631, 2646469360359, 8661172452084, 28242953648100, 91789599356325, 297398301914493, 960825283108362, 3095992578904722
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OFFSET
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3,2
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COMMENTS
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With offset = 2, a(n) is the number of length n words on alphabet {u,v,w,z} such that each word contains exactly 2 u's. - Zerinvary Lajos, Dec 29 2007
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LINKS
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FORMULA
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Numerators of sequence a[3,n] in (b^2)[i,j]) where b[i,j] = binomial(i-1, j-1)/2^(i-1) if j <= i, 0 if j > i.
a(n) = 3^(n-3)*binomial(n-1, 2).
G.f.: (x/(1-3*x))^3. (Third convolution of A000244, powers of 3.) (End)
The sequence 0, 1, 9, 54, ... has e.g.f.: (x + 3*x^2/2)*exp(3*x)/. - Paul Barry, Jul 23 2003
E.g.f.: E(0) where E(k) = 1 + 3*(2*k+3)*x/((2*k+1)^2 - 3*x*(k+2)*(2*k+1)^2/(3*x*(k+2) + 2*(k+1)^2/E(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 23 2012
Sum_{n>=3} 1/a(n) = 6 - 12*log(3/2).
Sum_{n>=3} (-1)^(n+1)/a(n) = 24*log(4/3) - 6. (End)
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MATHEMATICA
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nn=41; Drop[Range[0, nn]!CoefficientList[Series[Exp[x]^3 x^2/2!, {x, 0, nn}], x], 2] (* Geoffrey Critzer, Oct 03 2013 *)
LinearRecurrence[{9, -27, 27}, {1, 9, 54}, 40] (* G. C. Greubel, May 12 2021 *)
Abs[Take[CoefficientList[Series[1/(1+3x^2)^3, {x, 0, 60}], x], {1, -1, 2}]] (* Harvey P. Dale, Mar 03 2022 *)
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PROG
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(Sage) [3^(n-3)*binomial(n-1, 2) for n in range(3, 40)] # Zerinvary Lajos, Mar 10 2009
(Magma) [3^(n-3)*Binomial(n-1, 2): n in [3..40]]; // G. C. Greubel, May 12 2021
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CROSSREFS
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Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), this sequence (q=3), A038845 (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), A081139 (q=9), A081140 (q=10), A081141 (q=11), A081142 (q=12), A027476 (q=15).
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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