A138041 a(1) = 1, a(2) = 10; for n>2, a(n+1) = 4*a(n) + 6*a(n-1). Also a(n) = upper left term in the 2 X 2 matrix [1,3; 3,3].

1, 10, 46, 244, 1252, 6472, 33400, 172432, 890128, 4595104, 23721184, 122455360, 632148544, 3263326336, 16846196608, 86964744448, 448936157440, 2317533096448, 11963749330432, 61760195900416, 318823279584256

Offset

1,2

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (4, 6).

Formula

a(n)/a(n-1) tends to (2 + sqrt(10)) = 5.16227766... (a root of x^2 - 4*x - 6 and an eigenvalue of the matrix).

a(n) mod 9 == 1.

O.g.f.: -x*(1+6*x)/(-1+4*x+6*x^2). a(n) = A085939(n)+6*A085939(n-1). - R. J. Mathar, Mar 03 2008

From the characteristic polynomial of the matrix we get g.f.: (6*x + 1)/(-6*x^2 - 4*x + 1), with roots a=-(2+sqrt(10))/6, b=-(2-sqrt(10))/6. Let A=3+3*sqrt(10)/10 and B=3-3*sqrt(10)/10. Then a(n) = (A*(1/a)^n + B*(1/b)^n)/6. - Lambert Herrgesell (zero815(AT)googlemail.com), Apr 04 2008

Example

a(4) = 244 = 4*46 + 6*10 = 4*a(3) + 6*a(2).
a(4) = 244 = upper left term in [1,3; 3,3]^4.

Mathematica

a = {1, 10}; Do[AppendTo[a, 4*a[[ -1]] + 6*a[[ -2]]], {25}]; a (* Stefan Steinerberger *)
LinearRecurrence[{4, 6}, {1, 10}, 30] (* Harvey P. Dale, Mar 09 2014 *)

Crossrefs

Sequence in context: A115712 A199313 A003765 * A351901 A219597 A000832

Adjacent sequences: A138038 A138039 A138040 * A138042 A138043 A138044

Keyword

nonn

Author

Gary W. Adamson, Mar 02 2008

Extensions

More terms from Stefan Steinerberger and R. J. Mathar, Mar 02 2008

Definition corrected by Paolo P. Lava, Jun 03 2008

Status

approved