A164549 - OEIS

A164549

a(n) = 4*a(n-1) + 2*a(n-2) for n > 1; a(0) = 1, a(1) = 6.

1, 6, 26, 116, 516, 2296, 10216, 45456, 202256, 899936, 4004256, 17816896, 79276096, 352738176, 1569504896, 6983495936, 31072993536, 138258966016, 615181851136, 2737245336576, 12179345048576, 54191870867456

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OFFSET

0,2

COMMENTS

Binomial transform of A123011. Inverse binomial transform of A164550.

INVERT transform of the sequence (1, 5, 5*3, 5*3^2, 5*3^3, 5*3^4, ...); i.e., of (1, 5, 15, 45, 135, 405, ...). The sequence can also be obtained by extracting the upper left terms in matrix powers of [(1,5); (1,3)]. - Gary W. Adamson, Jul 31 2016

The sequence is A090017 (1, 4, 18, 80, 356, ...) convolved with (1, 2, 0, 0, 0, ...). Also, the upper left terms extracted from matrix powers of [(1,5); (1,3)]. - Gary W. Adamson, Aug 20 2016

LINKS

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (4,2).

FORMULA

a(n) = ((3+2*sqrt(6))*(2+sqrt(6))^n + (3-2*sqrt(6))*(2-sqrt(6))^n)/6.

G.f.: (1+2*x)/(1-4*x-2*x^2).

a(n) = (i*sqrt(2))^n*(ChebyshevU(n, -i*sqrt(2)) - sqrt(2)*i*ChebyshevU(n-1, -i*sqrt(2))). - G. C. Greubel, Jul 16 2021

MATHEMATICA

LinearRecurrence[{4, 2}, {1, 6}, 30] (* Harvey P. Dale, Mar 16 2013 *)

CoefficientList[Series[(1 +2x)/(1 -4x -2x^2), {x, 0, 24}], x] (* Michael De Vlieger, Aug 02 2016 *)

PROG

(Magma) [ n le 2 select 5*n-4 else 4*Self(n-1)+2*Self(n-2): n in [1..22] ];