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A215512 a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3), with a(0)=1, a(1)=3, a(2)=8.

%I A215512 #32 Sep 08 2022 08:46:03
%S A215512 1,3,8,23,70,220,703,2265,7327,23748,77043,250054,811760,2635519,
%T A215512 8557089,27784091,90213440,292919743,951102166,3088205812,10027335807,
%U A215512 32558546329,105716922615,343260670908,1114560365179,3618954723062,11750672095144,38154192502527
%N A215512 a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3), with a(0)=1, a(1)=3, a(2)=8.
%C A215512 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 A215512 sqrt(7)*b(n) = s(2)*c(4)^(2*n) + s(4)*c(1)^(2*n) + s(1)*c(2)^(2*n),
%C A215512 sqrt(7)*c(n) = s(4)*c(4)^(2*n) + s(1)*c(1)^(2*n) + s(2)*c(2)^(2*n), and
%C A215512 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 A215512 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 A215512 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 A215512 We note that a(n)=A000782(n) for every n=0,1,...,4 and A000782(5)-a(5)=2.
%C A215512 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 A215512 G. C. Greubel, <a href="/A215512/b215512.txt">Table of n, a(n) for n = 0..1000</a>
%H A215512 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 A215512 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5, -6, 1).
%F A215512 G.f.: (1-2*x-x^2)/(1-5*x+6*x^2-x^3).
%e A215512 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 A215512 LinearRecurrence[{5,-6,1}, {1,3,8}, 50]
%o A215512 (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 A215512 (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 A215512 Cf.A215694, A215695, A215007, A215008, A215143, A215493, A215494, A215510, A215575, A215455, A214683, A214699.
%K A215512 nonn,easy
%O A215512 0,2
%A A215512 _Roman Witula_, Aug 14 2012