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A094429 Given the 3 X 3 matrix M = [0 1 0 / 0 0 1 / 7 -14 7], a(n) = (-) rightmost term of M^n * [1 1 1]. 10

%I

%S 0,7,42,196,833,3381,13377,52136,201341,773122,2958032,11291903,

%T 43042727,163918671,623875840,2373568575,9028148962,34334213564,

%U 130560389505,496440779373,1887579497489,7176808297736,27286630574917

%N Given the 3 X 3 matrix M = [0 1 0 / 0 0 1 / 7 -14 7], a(n) = (-) rightmost term of M^n * [1 1 1].

%C M is derived from the Lucas polynomial: x^3 - 7*x^2 + 14*x - 7 with a root (and eigenvalue of the matrix): 3.801377358... = (2*sin(3*Pi/7))^2, the convergent of the sequence.

%C From _Roman Witula_, Sep 29 2012: (Start)

%C The Berndt-type sequence number 16 for the argument 2*Pi/7 (see Formula section and Crossrefs for other Berndt-type sequences for the argument 2*Pi/7 - for numbers from 1 to 18 without 16).

%C Note that all numbers of the form a(n)*7^(-1 - floor((n-1)/3)) are integers and even a(10) and a(11) are divisible by 7^5. (End)

%H G. C. Greubel, <a href="/A094429/b094429.txt">Table of n, a(n) for n = 1..1700</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 (7,-14,7).

%F From _Colin Barker_, Jun 19 2012: (Start)

%F a(n) = 7*a(n-1) - 14*a(n-2) + 7*a(n-3).

%F G.f.: 7*x^2*(1-x)/(1 - 7*x + 14*x^2 - 7*x^3). (End)

%F a(n) = c(4)*(s(1))^(2*n) + c(2)*(s(4))^(2*n) + c(1)*(s(2))^(2*n) = (-1/sqrt(7))*(c(1)*(s(1))^(2*n+3) + c(2)*(s(2))^(2*n+3) + c(3)*(s(3))^(2*n+3)) = (-1/sqrt(7))*(s(2)*(s(1))^(2*n+2) + s(4)*(s(2))^(2*n+2) + s(1)*(s(4))^(2*n+2)), where c(j) := 2*cos(2*Pi*j/7) and s(j) := 2*sin(2*Pi*j/7) (for the sums of the respective odd powers see A217274, see also A215493 and comments to A215494). For the proof of these formulas see Witula-Slota's paper. - _Roman Witula_, Jul 24 2012

%e a(5) = 833. M^5 * [1 1 1] = [ -42 -196 -833].

%e We have 4*a(4) - a(5) = 4*a(5) - a(6) = 7*a(2) = 49, 88*a(10) = 23*a(11), and a(3) = 6*a(2), which implies the equalities c(4)*(s(1))^6 + c(2)*(s(4))^6 + c(1)*(s(2))^6 = 6*(c(4)*(s(1))^4 + c(2)*(s(4))^4 + c(1)*(s(2))^4) and

%e s(2)*(s(1))^8 + s(4)*(s(2))^8 + s(1)*(s(4))^8 = 6*( s(2)*(s(1))^6 + s(4)*(s(2))^6 + s(1)*(s(4))^6). - _Roman Witula_, Sep 29 2012

%t Table[(MatrixPower[{{0, 1, 0}, {0, 0, 1}, {7, -14, 7}}, n].{-1, -1, -1})[[3]], {n, 23}] (* _Robert G. Wilson v_, May 08 2004 *)

%t LinearRecurrence[{7,-14,7}, {0,7,42}, 50] (* _Roman Witula_, Aug 13 2012 *)

%o (PARI) x='x+O('x^30); concat([0], Vec(7*x^2*(1-x)/(1-7*x+14*x^2-7*x^3))) \\ _G. C. Greubel_, May 09 2018

%o (PARI) a(n) = -(([0, 1, 0; 0, 0, 1; 7, -14, 7]^n)*[1,1,1]~)[3]; \\ _Michel Marcus_, May 10 2018

%o (MAGMA) I:=[0,7,42]; [n le 3 select I[n] else 7*Self(n-1) -14*Self(n-2) +7*Self(n-3): n in [1..30]]; // _G. C. Greubel_, May 09 2018

%Y Cf. A215007, A215008, A215143, A215493, A215494, A215510, A215512, A215575, A215694, A215695, A108716, A215794, A215828, A215817, A215877, A094430, A217274.

%K nonn,easy

%O 1,2

%A _Gary W. Adamson_, May 02 2004

%E More terms from _Robert G. Wilson v_, May 08 2004

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Last modified January 29 07:03 EST 2020. Contains 331337 sequences. (Running on oeis4.)