A216757 - OEIS

%I #28 Jan 29 2023 02:36:08

%S 0,-3,-15,-63,-252,-990,-3861,-15012,-58293,-226233,-877797,-3405564,

%T -13211910,-51254775,-198838152,-771371667,-2992450959,-11608875207,

%U -45035307612,-174709321686,-677764787229,-2629310751036,-10200109386213,-39570153919641,-153507871295037

%N a(n) = 6*a(n-1) - 9*a(n-2) + 3*a(n-3).

%C a(n) = X(2*n-1)/sqrt(3), where X(n) = 3*X(n-2) - sqrt(3)*X(n-3), with X(0)=3, X(1)=0, and X(2)=6.

%C The Berndt-type sequence number 13 for the argument 2*Pi/9 defined by the relation a(n)*sqrt(3) = s(1)^(2*n-1) - s(2)^(2*n-1) + s(4)^(2*n-1), where s(j) := 2*sin(2*Pi*j/9). For the respective sums with the even powers of sines - see A215634.

%C We note that X(n) = s(1)^n + (-s(2))^n + s(4)^n -- see Witula's book for details. Moreover the numbers of the form a(n)*3^(-1-floor((n-1)/3)) are integers.

%C The following summation formulas hold: Sum_{k=3..n} a(k) = 3*(2*a(n-1) - a(n-2) + 1), X(2*n+1) - X(1)*3^n = X(2*n+1) = -sqrt(3)*Sum_{k=1..n} X(2*(n-k))*3^(k-1), and X(2*n) - X(0)*3^n = X(2*n) - 3^(n+1) = -sqrt(3)*Sum{k=1..n} X(2*(n-k))*3^(k-1).

%D R. Witula, On some applications of formulas for sums of the unimodular complex numbers, Wyd. Pracowni Komputerowej Jacka Skalmierskiego, Gliwice 2011 (in Polish).

%H G. C. Greubel, <a href="/A216757/b216757.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-9,3).

%F a(n) = 6*a(n-1) - 9*a(n-2) + 3*a(n-3).

%F G.f.: -3*x^2*(1 - x)/(1 - 6*x + 9*x^2 - 3*x^3).

%F a(n) = Sum_{k=0..n} 3*(-1)^k*(binomial(2*n-1, n+9*k+7) - binomial(2*n-1, n+9*k+1)). - _Greg Dresden_, Jan 28 2023

%e We have s(1)^5 - s(2)^5 + s(4)^5 = 5*(s(1)^3 - s(2)^3 + s(4)^3) = -15*sqrt(3), s(1)^9 - s(2)^9 + s(4)^9 = 4*(s(1)^7 - s(2)^7 + s(4)^7) = -252*sqrt(3),

%e 39*(s(1)^11 - s(2)^11 + s(4)^11) = 10*(s(1)^13 - s(2)^13 + s(4)^13) = -38610*sqrt(3),

%e s(1)^7 - s(2)^7 + s(4)^7 = 4*(s(1)^5 - s(2)^5 + s(4)^5) + (s(1)^3 - s(2)^3 + s(4)^3) = -63*sqrt(3), and  s(1)^15 - s(2)^15 + s(4)^15 = 1000*(s(1)^5 - s(2)^5 + s(4)^5) + 4*(s(1)^3 - s(2)^3 + s(4)^3) = -15012*sqrt(3).

%e We note that a(6) = 3*(a(5) + a(4) + a(3)).

%t LinearRecurrence[{6,-9,3}, {0,-3,-15}, 30]

%t CoefficientList[Series[-3*x^2*(1 - x)/(1 - 6*x + 9*x^2 - 3*x^3), {x, 0,5 0}], x] (* _G. C. Greubel_, Apr 17 2017 *)

%o (PARI) concat(0,Vec(-3*(1-x)/(1-6*x+9*x^2-3*x^3)+O(x^99))) \\ _Charles R Greathouse IV_, Oct 01 2012

%Y Cf. A215634.

%K sign,easy

%O 1,2

%A _Roman Witula_, Sep 15 2012