|
|
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
|
|
|
1, 1, 3, 2, 6, 4, 12, 8, 24, 16, 48, 32, 96, 64, 192, 128, 384, 256, 768, 512, 1536, 1024, 3072, 2048, 6144, 4096, 12288, 8192, 24576, 16384, 49152, 32768, 98304, 65536, 196608, 131072, 393216, 262144, 786432, 524288, 1572864, 1048576
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
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).
The highest powers are given by the quarter-squares A002620(n-1). - Paul Barry, Mar 11 2007
|
|
LINKS
|
|
|
FORMULA
|
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)/(1-2*x^2); a(n)=2^floor(n/2)+2^((n-2)/2)*(1+(-1)^n)/2-0^n/2. - Paul Barry, Mar 11 2007
|
|
EXAMPLE
|
nu(0)=1;
nu(1)=1;
nu(2)=3;
nu(3)=5+2q;
nu(4)=11+8q+6q^2;
nu(5)=21+22q+20q^2+14q^3+4q^4;
nu(6)=43+60q+70q^2+64q^3+54q^4+28q^5+12q^6;
by listing the coefficients of the highest power in each nu(n), we get 1,1,3,2,6,4,12,...
|
|
MATHEMATICA
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|