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

Author

Gary W. Adamson, Mar 02 2008

Keywords

• nonn

Examples

			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.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (4, 6).

Programs

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 *)

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

Extensions

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

Definition corrected by Paolo P. Lava, Jun 03 2008