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A074324
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a(2n+1) = 3^n, a(2n) = 4*3^(n-1) except for a(0) = 1.
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5
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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
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OFFSET
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0,3
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COMMENTS
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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).
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LINKS
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FORMULA
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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).
a(n) = (1/6)*(7+(-1)^n)*3^floor(n/2) for n>0. - Ralf Stephan, Jul 19 2013
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EXAMPLE
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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,....
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MATHEMATICA
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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 *)
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PROG
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(Magma) [1] cat [(1/6)*(7+(-1)^n)*3^Floor(n/2):n in [1..40]]; // Vincenzo Librandi, Jul 20 2013
(PARI) a(n)=3^(n\2)\(3/4)^!bittest(n, 0) \\ M. F. Hasler, Dec 03 2014
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
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EXTENSIONS
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STATUS
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approved
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