A166552 -id:A166552 - cp's OEIS frontend

A162766 a(n) = 3*a(n-2) for n > 2; a(1) = 4, a(2) = 3.

4, 3, 12, 9, 36, 27, 108, 81, 324, 243, 972, 729, 2916, 2187, 8748, 6561, 26244, 19683, 78732, 59049, 236196, 177147, 708588, 531441, 2125764, 1594323, 6377292, 4782969, 19131876, 14348907, 57395628, 43046721, 172186884, 129140163

Offset: 1

Klaus Brockhaus, Jul 13 2009

nonn

Binomial transform is A162559. Second binomial transform is A077236.

Index entries for linear recurrences with constant coefficients, signature (0, 3).

Cf. A074324, A166552, A162559, A077236, A162436.

[ n le 2 select 5-n else 3*Self(n-2): n in [1..34] ];

a(n)=3^(n\2)*4^(n%2) \\ M. F. Hasler, Dec 03 2014

a(n) = (5-3*(-1)^n)*3^(1/4*(2*n-1+(-1)^n))/2.
G.f.: x*(4+3*x)/(1-3*x^2).
a(n) = A074324(n+1) = A166552(n+1) = 3^floor(n/2)*4^(n%2), where n%2 = 0 for n even, 1 for n odd. - M. F. Hasler, Dec 03 2014

G.f. corrected by Klaus Brockhaus, Sep 18 2009

A074324 a(2n+1) = 3^n, a(2n) = 4*3^(n-1) except for a(0) = 1.

Original entry on oeis.org

1, 1, 4, 3, 12, 9, 36, 27, 108, 81, 324, 243, 972, 729, 2916, 2187, 8748, 6561, 26244, 19683, 78732, 59049, 236196, 177147, 708588, 531441, 2125764, 1594323, 6377292, 4782969, 19131876, 14348907, 57395628, 43046721, 172186884, 129140163

Offset: 0

Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002

Also: Coefficient of the highest power of q in the expansion of nu(0)=1, nu(1)=b and for n>=2, nu(n)=b*nu(n-1)+lambda*(n-1)_q*nu(n-2) with (b,lambda)=(1,3), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity.
Instead of listing the coefficients of the highest power of q in each nu(n), if we list the coefficients of the smallest power of q (i.e., constant terms), we get a weighted Fibonacci numbers described by f(0)=1, f(1)=1, for n>=2, f(n)=f(n-1)+3f(n-2).
Sequences A162766, A166552 are essentially the same. - M. F. Hasler, Dec 03 2014

			nu(0)=1;
nu(1)=1;
nu(2)=4;
nu(3)=7+3q;
nu(4)=19+15q+12q^2;
nu(5)=40+45q+42q^2+30q^3+9q^4;
nu(6)=97+147q+180q^2+168q^3+147q^4+81q^5+36q^6;
by listing the coefficients of the highest power in each nu(n), we get, 1,1,4,3,12,9,36,....

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Beattie, S. Dăscălescu and S. Raianu, Lifting of Nichols Algebras of Type B_2, arXiv:math/0204075 [math.QA], 2002.
Index entries for linear recurrences with constant coefficients, signature (0,3).

Cf. A006130.

[1] cat [(1/6)*(7+(-1)^n)*3^Floor(n/2):n in [1..40]]; // Vincenzo Librandi, Jul 20 2013

CoefficientList[Series[-(1 + x + x^2) / (-1 + 3 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 20 2013 *)
LinearRecurrence[{0,3},{1,1,4},40] (* Harvey P. Dale, Mar 13 2016 *)

a(n)=3^(n\2)\(3/4)^!bittest(n,0) \\ M. F. Hasler, Dec 03 2014

For given b and lambda, the recurrence relation is given by; t(0)=1, t(1)=b, t(2)=b^2+lambda and for n>=3, t(n) = lambda*t(n-2).
G.f.: -(1+x+x^2)/(-1+3*x^2). - R. J. Mathar, Dec 05 2007
a(n) = 3*a(n-2) for n>2. - Ralf Stephan, Jul 19 2013
a(n) = (1/6)*(7+(-1)^n)*3^floor(n/2) for n>0. - Ralf Stephan, Jul 19 2013

More terms from R. J. Mathar, Dec 05 2007

Simpler definition from M. F. Hasler, Dec 03 2014

cp's OEIS Frontend

A162766 a(n) = 3*a(n-2) for n > 2; a(1) = 4, a(2) = 3.

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