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A095151
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a(n+3) = 3*a(n+2) - 2*a(n+1) + 1 with a(0)=0, a(1)=2.
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14
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0, 2, 7, 18, 41, 88, 183, 374, 757, 1524, 3059, 6130, 12273, 24560, 49135, 98286, 196589, 393196, 786411, 1572842, 3145705, 6291432, 12582887, 25165798, 50331621, 100663268, 201326563, 402653154, 805306337, 1610612704, 3221225439
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OFFSET
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0,2
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COMMENTS
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A sequence generated from a Bell difference row matrix, companion to A095150.
A095150 uses the same recursion rule but the multiplier [1 1 1] instead of [1 0 0].
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LINKS
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FORMULA
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Let M = a 3 X 3 matrix having Bell triangle difference terms (A095149 is composed of differences of the Bell triangle A011971): (fill in the 3 X 3 matrix with zeros): [1 0 0 / 1 1 0 / 2 1 2] = M. Then M^n * [1 0 0] = [1 n a(n)].
a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3).
G.f.: x*(2-x)/((1-x)^2*(1-2*x)). (End)
Let Prod_{i=0..n-1} (1+x^{2^i}+x^{2*2^i}) =
Sum_{j=0..d} b_j x^j, where d=2^{n+1}-2. Then
a(n)=Sum_{j=0..d-1} b_j/b_{j+1} (proved). - R. P. Stanley, Aug 27 2019
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EXAMPLE
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a(6) = 183 = 3*88 -2*41 + 1.
a(4) = 41 since M^4 * [1 0 0] = [1 4 41].
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MAPLE
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a[0]:=0:a[1]:=0:for n from 2 to 50 do a[n]:=2*a[n-1]+n od: seq(a[n], n=1..31); # Zerinvary Lajos, Feb 22 2008
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MATHEMATICA
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a[n_] := (MatrixPower[{{1, 0, 0}, {1, 1, 0}, {2, 1, 2}}, n].{{1}, {0}, {0}})[[3, 1]]; Table[ a[n], {n, 30}] (* Robert G. Wilson v, Jun 05 2004 *)
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PROG
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(PARI) vector(30, n, n--; 3*2^n -(n+3)) \\ G. C. Greubel, Jul 26 2019
(Magma) [3*2^n -(n+3): n in [0..30]]; // G. C. Greubel, Jul 26 2019
(Sage) [3*2^n -(n+3) for n in (0..30)] # G. C. Greubel, Jul 26 2019
(GAP) List([0..30], n-> 3*2^n -(n+3)); # G. C. Greubel, Jul 26 2019
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CROSSREFS
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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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