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%I #32 Sep 08 2022 08:46:03
%S 1,3,8,23,70,220,703,2265,7327,23748,77043,250054,811760,2635519,
%T 8557089,27784091,90213440,292919743,951102166,3088205812,10027335807,
%U 32558546329,105716922615,343260670908,1114560365179,3618954723062,11750672095144,38154192502527
%N a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3), with a(0)=1, a(1)=3, a(2)=8.
%C The Berndt-type sequence number 7 for the argument 2Pi/7 defined by the relation: sqrt(7)*a(n) = s(1)*c(4)^(2*n) + s(2)*c(1)^(2*n) + s(4)*c(2)^(2*n), where c(j):=2*cos(2*Pi*j/7) and s(j):=2*sin(2*Pi*j/7). If we additionally defined the following sequences:
%C sqrt(7)*b(n) = s(2)*c(4)^(2*n) + s(4)*c(1)^(2*n) + s(1)*c(2)^(2*n),
%C sqrt(7)*c(n) = s(4)*c(4)^(2*n) + s(1)*c(1)^(2*n) + s(2)*c(2)^(2*n), and
%C sqrt(7)*a1(n) = s(1)*c(4)^(2*n+1) + s(2)*c(1)^(2*n+1) + s(4)*c(2)^(2*n+1),
%C sqrt(7)*b1(n) = s(2)*c(4)^(2*n+1) + s(4)*c(1)^(2*n+1) + s(1)*c(2)^(2*n+1),
%C sqrt(7)*c1(n) = s(4)*c(4)^(2*n+1) + s(1)*c(1)^(2*n+1) + s(2)*c(2)^(2*n+1), then the following simple relationships between elements of these sequences hold true: a(n)=c1(n), c(n+1)=a1(n), -a(n)-b(n)=b1(n), which means that the sequences a1(n), b1(n), and c1(n) are completely and in very simple way determined by the sequences a(n), b(n) and c(n). However the last one's satisfy the following system of recurrence equations: a(n+1) = 2*a(n) + b(n), b(n+1) = a(n) + 2*b(n) - c(n), c(n+1) = c(n) - b(n). We have b(n)=A215694(n) and c(n)=A215695(n).
%C We note that a(n)=A000782(n) for every n=0,1,...,4 and A000782(5)-a(5)=2.
%C From general recurrence relation: a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3), i.e. a(n) = 5*(a(n-1)-a(n-2)) + (a(n-3)-a(n-2)) the following summation formula can be easily obtained: sum{k=3,..,n} a(k) = 5*a(n-1)-a(n-2)+a(0)-5*a(1). Hence in discussed sequence it follows that: sum{k=3,..,n} a(k) = 5*a(n-1) - a(n-2) - 14.
%H G. C. Greubel, <a href="/A215512/b215512.txt">Table of n, a(n) for n = 0..1000</a>
%H Roman Witula and Damian Slota, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Slota/witula13.html">New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7</a>, Journal of Integer Sequences, Vol. 10 (2007), Article 07.5.6
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5, -6, 1).
%F G.f.: (1-2*x-x^2)/(1-5*x+6*x^2-x^3).
%e We have a(6) = 10*a(4)+a(1), a(5) = 11*(a(3)-a(1)), a(10)-a(4)+a(3)+a(1)+a(0) = 77*10^3, and a(11)-a(4)+a(3)-a(2)+a(0) = 25*10^4 = (5^6)*(2^4).
%t LinearRecurrence[{5,-6,1}, {1,3,8}, 50]
%o (PARI) x='x+O('x^30); Vec((1-2*x-x^2)/(1-5*x+6*x^2-x^3)) \\ _G. C. Greubel_, Apr 23 2018
%o (Magma) I:=[1,3,8]; [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 23 2018
%Y Cf.A215694, A215695, A215007, A215008, A215143, A215493, A215494, A215510, A215575, A215455, A214683, A214699.
%K nonn,easy
%O 0,2
%A _Roman Witula_, Aug 14 2012
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