A232732 a(n) = Sum_{k=0..2*n} (-1)^k * binomial(12*n,6*k).

-922, 2434966, -6575675482, 17767242206806, -48007067114436442, 129715076631793674646, -350490088991612425827802, 947024090736392816800774486, -2558858742679396890519761433562, 6914035375695623821224314247122326, -18681721026270831871754901657845477722

Offset

1,1

Comments

a(2n) == 1 (mod 3); a(2n+1) == 2 (mod 3) Any proof?

This follows from the recurrence relation. - Charles R Greathouse IV, Dec 13 2013

Links

Colin Barker, Table of n, a(n) for n = 1..290

Index entries for linear recurrences with constant coefficients, signature (-2766,-172929,-64).

Formula

a(n) = (-1)^n/3 * (2^(6*n) + ((sqrt(3)+1)^(12*n) + (sqrt(3)-1)^(12*n))/2^(6*n)). - Vaclav Kotesovec, Dec 06 2013

G.f.: -2*x*(32*x^2+57643*x+461) / ((64*x+1)*(x^2+2702*x+1)). - Colin Barker, Dec 06 2013

a(n) = -2766*a(n-1)-172929*a(n-2)-64*a(n-3). - Wesley Ivan Hurt, Jan 15 2022

Maple

A232732:=n->add((-1)^i*binomial(12*n, 6*i), i=0..2*n); seq(A232732(n), n=1..20); # Wesley Ivan Hurt, Dec 06 2013

Mathematica

A[n_] := Sum[(-1)^k Binomial[12 n, 6 k], {k, 0, 2n}]; Array[A, 14]
CoefficientList[Series[-2 (32 x^2 + 57643 x + 461) / ((64 x + 1) (x^2 + 2702 x + 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 09 2014 *)
LinearRecurrence[{-2766, -172929, -64}, {-922, 2434966, -6575675482}, 20] (* Harvey P. Dale, Apr 18 2019 *)

Prog

(PARI) a(n)=sum(k=0, 2*n, (-1)^k*binomial(12*n, 6*k)) \\ Charles R Greathouse IV, Dec 13 2013
(PARI) Vec(-2*x*(32*x^2+57643*x+461)/((64*x+1)*(x^2+2702*x+1)) + O(x^100)) \\ Colin Barker, Nov 09 2014

Crossrefs

Cf. A232719.

Sequence in context: A289794 A235881 A051984 * A228673 A365469 A268849

Adjacent sequences: A232729 A232730 A232731 * A232733 A232734 A232735

Keyword

sign,easy

Author

José María Grau Ribas, Nov 29 2013

Status

approved