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A081568
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Third binomial transform of Fibonacci(n+1).
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8
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1, 4, 17, 75, 338, 1541, 7069, 32532, 149965, 691903, 3193706, 14745009, 68084297, 314394980, 1451837593, 6704518371, 30961415074, 142980203437, 660285858245, 3049218769908, 14081386948661, 65028302171639, 300302858766202
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OFFSET
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0,2
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COMMENTS
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Case k=3 of family of recurrences a(n) = (2k+1)*a(n-1) - A028387(k-1)*a(n-2) for n >= 2, with a(0) = 1 and a(1) = k + 1.
a(n) = 4^n*a(n;1/4) = Sum_{k=0..n} binomial(n,k) * (-1)^k * F(k-1) * 4^(n-k), which also implies Deléham's formula given below and where a(n;d), n = 0, 1, ..., d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also Witula's et al. papers). - Roman Witula, Jul 12 2012
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LINKS
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FORMULA
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a(n) = 7*a(n-1) - 11*a(n-2) for n >= 2, with a(0) = 1 and a(1) = 4.
a(n) = (1/2 - sqrt(5)/10)*(7/2 - sqrt(5)/2)^n + (sqrt(5)/10 + 1/2)*(sqrt(5)/2 + 7/2)^n = A099453(n) - 3*A099453(n-1).
G.f.: (1 - 3*x)/(1 - 7*x + 11*x^2).
G.f.: Q(0,u)/x - 1/x, where u = x/(1 - 3*x), Q(k,u) = 1 + u^2 + (k+2)*u - u*(k + 1 + u)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013
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MAPLE
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seq(coeff(series((1-3*x)/(1-7*x+11*x^2), x, n+1), x, n), n = 0 .. 30); # G. C. Greubel, Aug 12 2019
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MATHEMATICA
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CoefficientList[Series[(1-3x)/(1 -7x +11x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 09 2013 *)
LinearRecurrence[{7, -11}, {1, 4}, 30] (* Harvey P. Dale, Feb 01 2015 *)
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PROG
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(Magma) I:=[1, 4]; [n le 2 select I[n] else 7*Self(n-1)-11*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 09 2013
(PARI) Vec((1-3*x)/(1-7*x+11*x^2) + O(x^30)) \\ Altug Alkan, Dec 10 2015
(Sage)
P.<x> = PowerSeriesRing(ZZ, prec)
return P((1-3*x)/(1-7*x+11*x^2)).list()
(GAP) a:=[1, 4];; for n in [3..30] do a[n]:=7*a[n-1]-11*a[n-2]; od; a; # G. C. Greubel, Aug 12 2019
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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