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A104449
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Fibonacci sequence with initial values a(0) = 3 and a(1) = 1.
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13
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3, 1, 4, 5, 9, 14, 23, 37, 60, 97, 157, 254, 411, 665, 1076, 1741, 2817, 4558, 7375, 11933, 19308, 31241, 50549, 81790, 132339, 214129, 346468, 560597, 907065, 1467662, 2374727, 3842389, 6217116, 10059505, 16276621, 26336126, 42612747, 68948873, 111561620
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OFFSET
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0,1
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COMMENTS
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The old name was: The Pibonacci numbers (a Fibonacci-type sequence): each term is the sum of the two previous terms.
The 6th row in the Wythoff array begins with the 6th term of the sequence (14, 23, 37, 60, 97, 157, ...). a(n) = f(n-3) + f(n+2) for the Fibonacci numbers f(n) = f(n-1) + f(n-2); f(0) = 0, f(1) = 1.
(a(2*k), a(2*k+1)) give for k >= 0 the proper positive solutions of one of two families (or classes) of solutions (x, y) of the indefinite binary quadratic form x^2 + x*y - y^2 of discriminant 5 representing 11. The other family of such solutions is given by (x2, y2) = (b(2*k), b(2*k+1)) with b = A013655. See the formula in terms of Chebyshev S polynomials S(n, 3) = A001906(n+1) below, which follows from the fundamental solution (3, 1) by applying positive powers of the automorphic matrix given in a comment in A013655. See also A089270 with the Alfred Brousseau link with D = 11. - Wolfdieter Lang, May 28 2019
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REFERENCES
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V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.
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LINKS
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FORMULA
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a(n) = a(n-1) + a(n-2) with a(0) = 3, a(1) = 1.
a(n) = ( (3*sqrt(5)-1)*((1+sqrt(5))/2)^n + (3*sqrt(5)+1)*((1-sqrt(5) )/2)^n )/(2*sqrt(5)). - Bogart B. Strauss, Jul 19 2013
Bisection: a(2*k) = 4*S(k-1, 3) - 3*S(k-2, 3), a(2*k+1) = 2*S(k-1, 3) + S(k, 3) for k >= 0, with the Chebyshev S(n, 3) polynomials from A001906(n+1) for n >= -1. - Wolfdieter Lang, May 28 2019
a(3n + 4)/a(3n + 1) = continued fraction 4,4,4,...,4,9 (that's n 4's followed by a single 9). - Greg Dresden and Shaoxiong Yuan, Jul 16 2019
E.g.f.: (exp((1/2)*(1 - sqrt(5))*x)*(1 + 3*sqrt(5) + (- 1 + 3*sqrt(5))*exp(sqrt(5)*x)))/(2*sqrt(5)). - Stefano Spezia, Jul 18 2019
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MAPLE
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a:=n->3*fibonacci(n-1)+fibonacci(n): seq(a(n), n=0..40); # Zerinvary Lajos, Oct 05 2007
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MATHEMATICA
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LinearRecurrence[{1, 1}, {3, 1}, 40] (* Harvey P. Dale, May 23 2014 *)
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PROG
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(Magma) [Fibonacci(n-1) + Lucas(n): n in [0..40]]; // G. C. Greubel, May 29 2019
(Sage) ((3-2*x)/(1-x-x^2)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, May 29 2019
(GAP) a:=[3, 1];; for n in [3..40] do a[n]:=a[n-1]+a[n-2]; od; a; # G. C. Greubel, May 29 2019
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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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EXTENSIONS
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STATUS
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approved
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