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A129905
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G.f. (2*x+1)*(1-x)/((x+1)*(x^2-3*x+1)).
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1
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1, 3, 6, 17, 43, 114, 297, 779, 2038, 5337, 13971, 36578, 95761, 250707, 656358, 1718369, 4498747, 11777874, 30834873, 80726747, 211345366, 553309353, 1448582691, 3792438722, 9928733473, 25993761699, 68052551622, 178163893169
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OFFSET
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0,2
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COMMENTS
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a(n+2) - a(n) = A054486(n+1)
Form the infinite recursive array R(i,j) as follows: R(1,j) = F(j), R(2,j) = L(j) and for i>2, R(i,j) = R(i-1,j)+R(i-2,j) where F(j) is the j^th Fibonacci number and L(j) is the j^th Lucas number. Then for i>0, R(i,i) = a(i-1):
1 1 2 3 5 8 13 ...
1 3 4 7 11 18 29 ...
2 4 6 10 16 26 42 ...
3 7 10 17 27 44 71 ...
5 11 16 27 43 70 113 ...
8 18 26 44 70 114 184 ...
13 29 42 71 113 184 297 ...
..................................
- Andrew Rupinski, Jan. 31, 2011 -
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LINKS
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Table of n, a(n) for n=0..27.
Index entries for sequences related to linear recurrences with constant coefficients
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FORMULA
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a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3); a(n) = -1/5*(3/2+1/2*sqrt(5))^n*sqrt(5)+4/5*(3/2+1/2*sqrt(5))^n+1/5*(3/2-1/2*sqrt(5))^n*sqrt(5)+4/5*(3/2-1/2*sqrt(5))^n+2/5*(-1)^n
a(n) = -2*(-1)^n/5-8*A001906(n)/5+7*A001906(n+1)/5. [From R. J. Mathar, Nov 10 2009]
a(n)= (Fibonacci(n-2)^2+Fibonacci(n+2)^2+Fibonacci(2n))/2, [From Gary Detlefs Dec 20 2010]
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MATHEMATICA
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CoefficientList[ Series[(1 + x - 2 x^2)/(1 - 2 x - 2 x^2 + x^3), {x, 0, 27}], x] (* Or *)
t[1, k_] := Fibonacci@ k; t[2, k_] := LucasL@ k; t[n_, k_] := t[n, k] = t[n - 1, k] + t[n - 2, k]; Table[ t[n, n], {n, 28}] (* Robert G. Wilson v *)
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PROG
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Floretion Algebra Multiplication Program, FAMP Code: tesseq[A*B] with A = + .5'i + .5'j + .5'k + 'ji' + .5e ; B = + .5i' + .5j' + .5k' + 'ij' + .5e (apart from initial term)
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CROSSREFS
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Cf. A001906, A054486.
Sequence in context: A089264 A121399 A212421 * A143363 A216878 A006081
Adjacent sequences: A129902 A129903 A129904 * A129906 A129907 A129908
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KEYWORD
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easy,nonn
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AUTHOR
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Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Jun 04 2007
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STATUS
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approved
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