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A215694
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a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3) with a(0)=1, a(1)=2, a(2)=7.
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10
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1, 2, 7, 24, 80, 263, 859, 2797, 9094, 29547, 95968, 311652, 1011999, 3286051, 10669913, 34645258, 112492863, 365262680, 1186001480, 3850924183, 12503874715, 40599829957, 131826825678, 428039023363, 1389833992704, 4512762649020, 14652848312239, 47577499659779, 154483171074481, 501603705725970, 1628697001842743
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OFFSET
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0,2
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COMMENTS
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The Berndt-type sequence number 9 for the argument 2Pi/7 defined by the first trigonometric relation from section "Formula". For more connections with another sequences of trigonometric nature see comments to A215512 (a(n) is equal to the sequence b(n) in these comments) and Witula-Slota's reference (Section 3). We note that a(n)=A109682(n) for n=1,2,3,4. Moreover the following summation formula hold true: sum{k=3,..,n} a(k) = 5*a(n-1) - a(n-2) - 9, for every n=3,4,... - see comments to A215512.
The inverse binomial transform is 1,1, 4, 8, 19, 42, 95,... essentially a shifted, unsigned variant of A215112. - R. J. Mathar, Aug 22 2012
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LINKS
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FORMULA
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sqrt(7)*a(n) = s(4)*c(1)^(2*n) + s(1)*c(2)^(2*n) + s(2)*c(4)^(2*n), where c(j):=2*cos(2*Pi*j/7) and s(j):=2*sin(2*Pi*j/7).
G.f.: (1-3*x+3*x^2)/(1-5*x+6*x^2-x^3).
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EXAMPLE
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We have 10*a(3) = 3*a(4), a(0)+a(1)+3*a(2) = a(3), a(0)+a(2)+3*a(3) = a(4), a(1)+3*a(2)+3*a(4) = a(5), and a(6) = 3*a(5)+3*a(4)-a(1).
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MATHEMATICA
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LinearRecurrence[{5, -6, 1}, {1, 2, 7}, 50]
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PROG
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(Magma) I:=[1, 2, 7]; [n le 3 select I[n] else 5*Self(n-1) - 6*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Apr 25 2018
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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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STATUS
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approved
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