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A153882
a(n) = ((6 + sqrt(5))^n - (6 - sqrt(5))^n)/(2*sqrt(5)).
1
1, 12, 113, 984, 8305, 69156, 572417, 4725168, 38957089, 321004860, 2644388561, 21781512072, 179402099473, 1477598319444, 12169714749665, 100231029093216, 825511191878977, 6798972400658028, 55996821859648049, 461193717895377720
OFFSET
1,2
COMMENTS
Fourth binomial transform of A048877.
First differences are in A163146.
lim_{n -> infinity} a(n)/a(n-1) = 6 + sqrt(5) = 8.236067977499789696....
REFERENCES
S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
FORMULA
From Philippe Deléham, Jan 03 2009: (Start)
a(n) = 12*a(n-1) - 31*a(n-2) for n>1, with a(0)=0, a(1)=1.
G.f.: x/(1 - 12*x + 31*x^2). (End)
a(n) = 12*a(n-1) - 31*a(n-2). - G. C. Greubel, Aug 31 2016
MATHEMATICA
LinearRecurrence[{12, -31}, {1, 12}, 25] (* or *) Table[((6 + sqrt(5))^n - (6 - sqrt(5))^n)/(2*sqrt(5)) , {n, 0, 25}] (* G. C. Greubel, Aug 31 2016 *)
PROG
(Magma) Z<x>:= PolynomialRing(Integers()); N<r>:=NumberField(x^2-5); S:=[ ((6+r)^n-(6-r)^n)/(2*r): n in [1..20] ]; [ Integers()!S[j]: j in [1..#S] ]; # Klaus Brockhaus, Jan 04 2009
(Sage) [lucas_number1(n, 12, 31) for n in range(1, 21)] # Zerinvary Lajos, Apr 27 2009
(Magma) I:=[1, 12]; [n le 2 select I[n] else 12*Self(n-1)-31*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 01 2016
CROSSREFS
Cf. A002163 (decimal expansion of sqrt(5)), A048877, A163146.
Sequence in context: A221368 A083767 A285154 * A357650 A322649 A198375
KEYWORD
nonn
AUTHOR
Al Hakanson (hawkuu(AT)gmail.com), Jan 03 2009
EXTENSIONS
Extended beyond a(7) by Klaus Brockhaus, Jan 04 2009
Edited by Klaus Brockhaus, Oct 11 2009
STATUS
approved