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%I #81 Sep 08 2022 08:45:10
%S 1,1,6,16,56,176,576,1856,6016,19456,62976,203776,659456,2134016,
%T 6905856,22347776,72318976,234029056,757334016,2450784256,7930904576,
%U 25664946176,83053510656,268766806016,869747654656,2814562533376,9108115685376,29474481504256
%N a(n) = 2*a(n-1) + 4*a(n-2), a(0)=1, a(1)=1.
%C Inverse binomial transform of A001077. Binomial transform of expansion of cosh(sqrt(5)*x) (1,0,5,0,25,...).
%C The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 5 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(5). - _Cino Hilliard_, Sep 25 2005
%C Numerators of fractions in the approximation of the square root of 5 satisfying: a(n) = (a(n-1)+c)/(a(n-1)+1), with c=5 and a(1)=1. For denominators see A063727. - _Mark Dols_, Jul 24 2009
%C Equals right border of triangle A143969. (1, 6, 16, 56, ...) = row sums of triangle A143969 and INVERT transform of (1, 5, 5, 5, ...). - _Gary W. Adamson_, Sep 06 2008
%C a(n) is the number of compositions of n when there are 1 type of 1 and 5 types of other natural numbers. - _Milan Janjic_, Aug 13 2010
%C From _Gary W. Adamson_, Jul 30 2016: (Start)
%C The sequence is case N=1 in an infinite set obtained by taking powers of the 2 X 2 matrix M = [(1,5); (1,N)], then extracting the upper left terms. The infinite set begins:
%C N=1 (A084057): 1, 6, 16, 56, 176, 576, 1856, ...
%C N=2 (A108306): 1, 6, 21, 81, 306, 1161, 4401, ...
%C N=3 (A164549): 1, 6, 26, 116, 516, 2296, 10216, ...
%C N=4 (A015449): 1, 6, 31, 161, 836, 4341, 22541, ...
%C N=5 (A000400): 1, 6, 36, 216, 1296, 7776, 46656, ...
%C N=6 (A049685): 1, 6, 41, 281, 1926, 13201, 90481, ...
%C N=7 (.......): 1, 6, 46, 356, 2756, 21336, 222712, ...
%C ...
%C Sequences in the above set can be obtained by taking INVERT transforms of the following:
%C N=1 INVERT transform of (1, 5, 5, 5, 5, 5, ...
%C N=2 ..."......"......". (1, 5, 10, 20, 40, 80, ...
%C N=3 ..."......"......". (1, 5, 15, 45, 135, 405, ...
%C N=4 ..."......"......". (1, 5, 20, 80, 320, 1280, ...
%C ...
%C with the pattern (1, 5, N*5, (N^2)*5, (N^3)*5, ...
%C It appears that the sequence generated from powers (n>0) of the matrix P = [(1,a); (1,b)], (a,b > 0), then extracting the upper left terms, is equal to the INVERT transform of the sequence starting: (1, a, b*a, (b^2)*a, (b^3)*a, ...). (End)
%D John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.
%H Vincenzo Librandi, <a href="/A084057/b084057.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,4).
%F a(n) = ((1+sqrt(5))^n + (1-sqrt(5))^n)/2.
%F G.f.: (1-x) / (1-2*x-4*x^2).
%F E.g.f.: exp(x) * cosh(sqrt(5)*x).
%F a(2n+1) = 2*a(n)*a(n+1) - (-4)^n. - Mario Catalani (mario.catalani(AT)unito.it), Jun 13 2003
%F a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*5^k . - _Paul Barry_, Jul 25 2004
%F a(n) = Sum_{k=0..n} A098158(n,k)*5^(n-k). - _Philippe Deléham_, Dec 26 2007
%F a(n) = 2^(n-1)*A000032(n). - _Mark Dols_, Jul 24 2009
%F If p(1)=1, and p(i)=5 for i>1, and if A is the Hessenberg matrix of order n defined by: A(i,j) = p(j-i+1) for i<=j, A(i,j):=-1, (i=j+1), and A(i,j):=0 otherwise, then, for n>=1, a(n)=det A. - _Milan Janjic_, Apr 29 2010
%F G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(5*k-1)/(x*(5*k+4) - 1/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, May 26 2013
%F a(n) = A063727(n) - A063272(n-1). - _R. J. Mathar_, Jun 06 2019
%F a(n) = 1 + 5*A014335(n). - _R. J. Mathar_, Jun 06 2019
%F Sum_{n>=1} 1/a(n) = A269992. - _Amiram Eldar_, Feb 01 2021
%t f[n_] := Simplify[((1 + Sqrt[5])^n + (1 - Sqrt[5])^n)/2]; Array[f, 28, 0] (* Or *)
%t LinearRecurrence[{2, 4}, {1, 1}, 28] (* _Robert G. Wilson v_, Sep 18 2013 *)
%t RecurrenceTable[{a[1] == 1, a[2] == 1, a[n] == 2 a[n-1] + 4 a[n-2]}, a, {n, 30}] (* _Vincenzo Librandi_, Jul 31 2016 *)
%t Table[2^(n-1) LucasL[n], {n, 0, 20}] (* _Vladimir Reshetnikov_, Sep 19 2016 *)
%o (Sage) from sage.combinat.sloane_functions import recur_gen2b; it = recur_gen2b(1,1,2,4, lambda n: 0); [next(it) for i in range(1,26)] # _Zerinvary Lajos_, Jul 09 2008
%o (Sage) [lucas_number2(n,2,-4)/2 for n in range(0, 26)] # _Zerinvary Lajos_, Apr 30 2009
%o (PARI) lucas(n)=fibonacci(n-1)+fibonacci(n+1)
%o a(n)=lucas(n)/2*2^n \\ _Charles R Greathouse IV_, Sep 18 2013
%o (Magma) I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1)+4*Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Jul 31 2016
%Y Cf. A046717, A002533, A098158, A143969, A269992.
%Y a(n) = A087131(n)/2.
%Y The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
%Y Cf. A108306, A164549, A015449, A000400, A049685
%K easy,nonn
%O 0,3
%A _Paul Barry_, May 10 2003
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