A138229 Expansion of (1-x)/(1-2x+6x^2).

1, 1, -4, -14, -4, 76, 176, -104, -1264, -1904, 3776, 18976, 15296, -83264, -258304, -17024, 1515776, 3133696, -2827264, -24456704, -31949824, 82840576, 357380096, 217716736, -1708847104, -4723994624, 805093376

Offset: 0

Author

Paul Barry, Mar 06 2008

Keywords

• easy
• sign

Comments

Binomial transform of [1, 0, -5, 0, 25, 0, -125, 0, 625, 0, ...]=: powers of -5 with interpolated zeros. - Philippe Deléham, Dec 02 2008

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..2571
• Beata Bajorska-Harapińska, Barbara Smoleń, Roman Wituła, On Quaternion Equivalents for Quasi-Fibonacci Numbers, Shortly Quaternaccis, Advances in Applied Clifford Algebras (2019) Vol. 29, 54.
• Index entries for linear recurrences with constant coefficients, signature (2,-6).

Crossrefs

Cf. A088139.

Programs

Mathematica

CoefficientList[Series[(1-x)/(1-2x+6x^2),{x,0,30}],x] (* or *) LinearRecurrence[ {2,-6},{1,1},30] (* Harvey P. Dale, Feb 29 2012 *)
TrigExpand@Table[6^(n/2) Cos[n ArcTan[Sqrt[5]]], {n, 0, 20}] (* or *)
Table[Sum[(-5)^k Binomial[n, 2 k], {k, 0, n/2}], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 20 2016 *)

Sage

[lucas_number2(n,2,6)/2 for n in range(0,28)] # Zerinvary Lajos, Jul 08 2008

Formula

From Philippe Deléham, Nov 14 2008: (Start)

a(n) = 2*a(n-1) - 6*a(n-2), a(0)=1, a(1)=1.

a(n) = Sum_{k=0..n} A098158(n,k)*(-5)^(n-k). (End)

a(n) = Sum_{k=0..n} A124182(n,k)*(-6)^(n-k). - Philippe Deléham, Nov 15 2008

G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(5*k+1)/(x*(5*k+6) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013

a(n) = real part of the quaternion (1 + i + 2*j)^n. - Peter Bala, Mar 29 2015