A084057 - OEIS

A084057

a(n) = 2*a(n-1) + 4*a(n-2), a(0)=1, a(1)=1.

1, 1, 6, 16, 56, 176, 576, 1856, 6016, 19456, 62976, 203776, 659456, 2134016, 6905856, 22347776, 72318976, 234029056, 757334016, 2450784256, 7930904576, 25664946176, 83053510656, 268766806016, 869747654656, 2814562533376, 9108115685376, 29474481504256 (list; graph; refs; listen; history; text; internal format)

	OFFSET	`0,3`
	COMMENTS	`Inverse binomial transform of A001077. Binomial transform of expansion of cosh(sqrt(5)x) (1,0,5,0,25,...).` `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` `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` `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` `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` `From Gary W. Adamson, Jul 30 2016: (Start)` `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:` `N=1 (A084057): 1, 6, 16, 56, 176, 576, 1856, ...` `N=2 (A108306): 1, 6, 21, 81, 306, 1161, 4401, ...` `N=3 (A164549): 1, 6, 26, 116, 516, 2296, 10216, ...` `N=4 (A015449): 1, 6, 31, 161, 836, 4341, 22541, ...` `N=5 (A000400): 1, 6, 36, 216, 1296, 7776, 46656, ...` `N=6 (A049685): 1, 6, 41, 281, 1926, 13201, 90481, ...` `N=7 (.......): 1, 6, 46, 356, 2756, 21336, 222712, ...` `...` `Sequences in the above set can be obtained by taking INVERT transforms of the following:` `N=1 INVERT transform of (1, 5, 5, 5, 5, 5, ...` `N=2 ..."......"......". (1, 5, 10, 20, 40, 80, ...` `N=3 ..."......"......". (1, 5, 15, 45, 135, 405, ...` `N=4 ..."......"......". (1, 5, 20, 80, 320, 1280, ...` `...` `with the pattern (1, 5, N5, (N^2)5, (N^3)5, ...` `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, ba, (b^2)a, (b^3)*a, ...). (End)`
	REFERENCES	`John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.`
	LINKS	`Vincenzo Librandi, Table of n, a(n) for n = 0..1000` `Index entries for linear recurrences with constant coefficients, signature (2,4).`
	FORMULA	`a(n) = ((1+sqrt(5))^n + (1-sqrt(5))^n)/2.` `G.f.: (1-x) / (1-2x-4x^2).` `E.g.f.: exp(x) * cosh(sqrt(5)x).` `a(2n+1) = 2a(n)a(n+1) - (-4)^n. - Mario Catalani (mario.catalani(AT)unito.it), Jun 13 2003` `a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)5^k . - Paul Barry, Jul 25 2004` `a(n) = Sum_{k=0..n} A098158(n,k)5^(n-k). - Philippe Deléham, Dec 26 2007` `a(n) = 2^(n-1)A000032(n). - Mark Dols, Jul 24 2009` `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` `G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x(5k-1)/(x(5k+4) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013` `a(n) = A063727(n) - A063272(n-1). - R. J. Mathar, Jun 06 2019` `a(n) = 1 + 5A014335(n). - R. J. Mathar, Jun 06 2019` `Sum_{n>=1} 1/a(n) = A269992. - Amiram Eldar, Feb 01 2021`
	MATHEMATICA	`f[n_] := Simplify[((1 + Sqrt[5])^n + (1 - Sqrt[5])^n)/2]; Array[f, 28, 0] (* Or )` `LinearRecurrence[{2, 4}, {1, 1}, 28] ( Robert G. Wilson v, Sep 18 2013 )` `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 )` `Table[2^(n-1) LucasL[n], {n, 0, 20}] ( Vladimir Reshetnikov, Sep 19 2016 *)`
	PROG	`(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` `(Sage) [lucas_number2(n, 2, -4)/2 for n in range(0, 26)] # Zerinvary Lajos, Apr 30 2009` `(PARI) lucas(n)=fibonacci(n-1)+fibonacci(n+1)` `a(n)=lucas(n)/22^n \\ Charles R Greathouse IV, Sep 18 2013` `(Magma) I:=[1, 1]; [n le 2 select I[n] else 2Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 31 2016`
	CROSSREFS	`Cf. A046717, A002533, A098158, A143969, A269992.` `a(n) = A087131(n)/2.` `The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.` `Cf. A108306, A164549, A015449, A000400, A049685` `Sequence in context: A316984 A192000 A032282 * A163302 A223028 A091649` `Adjacent sequences: A084054 A084055 A084056 * A084058 A084059 A084060`
	KEYWORD	`easy,nonn`
	AUTHOR	`Paul Barry, May 10 2003`
	STATUS	`approved`