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A074323 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)=(1,2), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity. 7

%I #30 Sep 22 2017 12:31:51

%S 1,1,3,2,6,4,12,8,24,16,48,32,96,64,192,128,384,256,768,512,1536,1024,

%T 3072,2048,6144,4096,12288,8192,24576,16384,49152,32768,98304,65536,

%U 196608,131072,393216,262144,786432,524288,1572864,1048576

%N 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)=(1,2), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity.

%C 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)+2f(n-2).

%C The highest powers are given by the quarter-squares A002620(n-1). - _Paul Barry_, Mar 11 2007

%H M. Beattie, S. Dăscălescu and S. Raianu, <a href="https://arxiv.org/abs/math/0204075">Lifting of Nichols Algebras of Type B_2</a>, arXiv:math/0204075 [math.QA], 2002.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,2).

%F 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).

%F G.f.: (1+x+x^2)/(1-2*x^2); a(n)=2^floor(n/2)+2^((n-2)/2)*(1+(-1)^n)/2-0^n/2. - _Paul Barry_, Mar 11 2007

%F a(0)=0, a(2n+1) = A000079, a(2n+2) = 3a(2n+1). a(2n)-a(2n+1) = A131577. - _Paul Curtz_, Mar 05 2008

%F a(2n+1) = 2^n = A000079(n), a(2n+2) = 3*A000079(n). Also a(2n)-a(2n+1) = A131577. a(2n+1)-a(2n)=2^n for n>0. - _Paul Curtz_, Apr 09 2008

%F a(n+1) = A010684(n)*A016116(n). - _R. J. Mathar_, Jul 08 2009

%e nu(0)=1;

%e nu(1)=1;

%e nu(2)=3;

%e nu(3)=5+2q;

%e nu(4)=11+8q+6q^2;

%e nu(5)=21+22q+20q^2+14q^3+4q^4;

%e nu(6)=43+60q+70q^2+64q^3+54q^4+28q^5+12q^6;

%e by listing the coefficients of the highest power in each nu(n), we get 1,1,3,2,6,4,12,...

%t Join[{1}, LinearRecurrence[{0, 2}, {1, 3}, 41]] (* _Jean-François Alcover_, Sep 22 2017 *)

%Y Cf. A001045.

%K nonn,easy

%O 0,3

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

%E More terms from _Paul Barry_, Mar 11 2007

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Last modified April 24 10:11 EDT 2024. Contains 371935 sequences. (Running on oeis4.)