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A215695
a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3) with a(0)=1, a(1)=0, a(2)=-2.
10
1, 0, -2, -9, -33, -113, -376, -1235, -4032, -13126, -42673, -138641, -450293, -1462292, -4748343, -15418256, -50063514, -162556377, -527819057, -1713820537, -5564744720, -18068619435, -58668449392, -190495275070, -618534298433, -2008368291137, -6521130940157, -21173979252396, -68751478912175, -223234649986656, -724838355712626
OFFSET
0,3
COMMENTS
The Berndt-type sequence number 10 for the argument 2Pi/7 defined by the first trigonometric relation from section "Formula". For additional informations and particularly connections with another sequences of trigonometric nature - see comments to A215512 (a(n) is equal to the sequence c(n) in these comments) and Witula-Slota's reference (Section 3).
The following summation formula hold true (see comments to A215512): Sum{k=3,..,n} a(k) = 5*a(n-1) - a(n-2) + 1, n=3,4,...
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(1)*c(1)^(2*n) + s(2)*c(2)^(2*n) + s(4)*c(4)^(2*n), where c(j):=2*cos(2*Pi*j/7) and s(j):=2*sin(2*Pi*j/7).
G.f.: (1-5*x+4*x^2)/(1-5*x+6*x^2-x^3).
a(n) = A005021(n) - 5*A005021(n-1) + 4*A005021(n-2). - R. J. Mathar, Aug 22 2012
EXAMPLE
We have a(8)=3*a(7)+3*a(5)-6*a(2) and a(9)=3*a(8)+3*a(6)-3*a(4)-a(1).
MATHEMATICA
LinearRecurrence[{5, -6, 1}, {1, 0, -2}, 50]
PROG
(PARI) x='x+O('x^30); Vec((1-5*x+4*x^2)/(1-5*x+6*x^2-x^3)) \\ G. C. Greubel, Apr 25 2018
(Magma) I:=[1, 0, -2]; [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
Cf. A215512 (the inverse binomial transform, up to signs), A215694.
Sequence in context: A122097 A073400 A048498 * A289600 A202206 A150921
KEYWORD
sign,easy
AUTHOR
Roman Witula, Aug 21 2012
STATUS
approved