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A171516
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a(n) = a(n-1) + a(n-2) + k, n>1; with a(0) = 1, a(1) = 2, k = 3.
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4
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1, 2, 6, 11, 20, 34, 57, 94, 154, 251, 408, 662, 1073, 1738, 2814, 4555, 7372, 11930, 19305, 31238, 50546, 81787, 132336, 214126, 346465, 560594, 907062, 1467659, 2374724, 3842386, 6217113, 10059502, 16276618, 26336123, 42612744, 68948870
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OFFSET
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0,2
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COMMENTS
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In an infinite set of sequences converging to phi, H(n+1) = H(n) + H(n-1) + k.
The coincident formula = H(n) = (H(n+1) + H(n-2))/2, then proof of convergence to phi follows, [Gatta and D'Amico]: To get H(4) such that the average of H(4) and H(1) = H(3), the authors write H(4) = 2H(3) - H1 = 2H(1) + 2H(2) + 2k - H(1) = H(2) + (H(1) + H(2) + k) + k = H(2) + H(3) + k, then applying the iterative process to the latter, H(n+1) = H(n) + H(n-1) + k.
Cf. A014739 for a(0) = 1, a(1) = 2, k = 2, getting:
A014739 = (1, 2, 5,. 9, 16, 27, 45, 74, 121, 197,...)
A171516 = (1, 2, 6, 11, 20, 34, 57, 94, 154, 251,...), we obtain
A000071 = (0, 0, 1, .2,..4,..7,.12,.20,..33,..54,...).
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LINKS
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FORMULA
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a(n) = a(n-1) + a(n-2) + 3, n>1; with a(0) = 1, a(1) = 2.
a(n) = 2*a(n-1) - a(n-3) = A000285(n+1) - 3.
G.f.: (1+2*x^2) / ((1-x)*(1-x-x^2)). (End)
a(n) = Fibonacci(n+3) + 2*Fibonacci(n+1) - 3. - G. C. Greubel, Jul 12 2019
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EXAMPLE
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a(5) = 34 = a(4) + a(3) = 3 = 34.
a(4) = 20 = (a(5) + a(2) / 2 = (34 + 6) / 2.
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MATHEMATICA
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LinearRecurrence[{2, 0, -1}, {1, 2, 6}, 40] (* Harvey P. Dale, Apr 07 2012 *)
With[{F=Fibonacci}, Table[F[n+3]+2*F[n+1]-3, {n, 0, 40}]] (* G. C. Greubel, Jul 12 2019 *)
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PROG
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(PARI) vector(40, n, n--; f=fibonacci; f(n+2)+2*f(n+1)-3) \\ G. C. Greubel, Jul 12 2019
(Magma) F:=Fibonacci; [F(n+3)+2*F(n+1)-3: n in [0..40]]; // G. C. Greubel, Jul 12 2019
(Sage) f=fibonacci; [f(n+2)+2*f(n+1)-3 for n in (0..40)] # G. C. Greubel, Jul 12 2019
(GAP) F:=Fibonacci;; List([0..40], n-> F(n+3)+2*F(n+1)-3); # G. C. Greubel, Jul 12 2019
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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