A215695 a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3) with a(0)=1, a(1)=0, a(2)=-2.
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
Examples
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).
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- 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
- Index entries for linear recurrences with constant coefficients, signature (5,-6,1).
Programs
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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
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Mathematica
LinearRecurrence[{5,-6,1}, {1,0,-2}, 50]
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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
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).
Comments