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A036216
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Expansion of 1/(1 - 3*x)^4; 4-fold convolution of A000244 (powers of 3).
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26
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1, 12, 90, 540, 2835, 13608, 61236, 262440, 1082565, 4330260, 16888014, 64481508, 241805655, 892820880, 3252418920, 11708708112, 41712272649, 147219785820, 515269250370, 1789882659180, 6175095174171, 21171754882872
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OFFSET
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0,2
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COMMENTS
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With three leading zeros, 3rd binomial transform of (0,0,0,1,0,0,0,0,...). - Paul Barry, Mar 07 2003
Number of n-permutations (n=4) of 4 objects u, v, w, z, with repetition allowed, containing exactly three u's. - Zerinvary Lajos, May 23 2008
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LINKS
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FORMULA
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a(n) = 3^n*binomial(n+3, 3).
G.f.: 1/(1 - 3*x)^4.
With three leading zeros, a(n) = 12*a(n-1) - 54*a(n-2) + 108*a(n-3) - 81*a(n-4), a(0) = a(1) = a(2) = 0, a(3) = 1. - Paul Barry, Mar 07 2003
With three leading zeros, C(n, 3)*3^(n-3) is the second binomial transform of C(n, 3). - Paul Barry, Jul 24 2003
Sum_{n>=0} 1/a(n) = 36*log(3/2) - 27/2.
Sum_{n>=0} (-1)^n/a(n) = 144*log(4/3) - 81/2. (End)
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MAPLE
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MATHEMATICA
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CoefficientList[Series[1/(1-3x)^4, {x, 0, 30}], x] (* or *) LinearRecurrence[ {12, -54, 108, -81}, {1, 12, 90, 540}, 30] (* Harvey P. Dale, Jul 27 2017 *)
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PROG
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(Sage) [3^n*binomial(n+3, 3) for n in range(30)] # Zerinvary Lajos, Mar 10 2009
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CROSSREFS
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Sequences of the form 3^n*binomial(n+m, m): A000244 (m=0), A027471 (m=1), A027472 (m=2), this sequence (m=3), A036217 (m=4), A036219 (m=5), A036220 (m=6), A036221 (m=7), A036222 (m=8), A036223 (m=9), A172362 (m=10).
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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