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A215694
a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3) with a(0)=1, a(1)=2, a(2)=7.
10
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
OFFSET
0,2
COMMENTS
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
LINKS
Roman Witula and Damian Slota, New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7, Journal of Integer Sequences, Vol. 10 (2007), Article 07.5.6
FORMULA
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).
a(n) = A005021(n)-3*A005021(n-1)+3*A005021(n-2). - R. J. Mathar, Aug 22 2012
EXAMPLE
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).
MATHEMATICA
LinearRecurrence[{5, -6, 1}, {1, 2, 7}, 50]
PROG
(PARI) Vec((1-3*x+3*x^2)/(1-5*x+6*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012
(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
CROSSREFS
Sequence in context: A027122 A132788 A109682 * A027124 A038765 A027126
KEYWORD
nonn,easy
AUTHOR
Roman Witula, Aug 21 2012
STATUS
approved