

A007482


a(n) is the number of subsequences of [ 1, ..., 2n ] in which each odd number has an even neighbor.
(Formerly M2893)


44



1, 3, 11, 39, 139, 495, 1763, 6279, 22363, 79647, 283667, 1010295, 3598219, 12815247, 45642179, 162557031, 578955451, 2061980415, 7343852147, 26155517271, 93154256107, 331773802863, 1181629920803, 4208437368135
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OFFSET

0,2


COMMENTS

The even neighbor must differ from the odd number by exactly one.
If we defined this sequence by the recurrence (a(n) = 3*a(n1) + 2*a(n2)) that it satisfies, we could prefix it with an initial 0.
a(n) equals term (1,2) in M^n, M = the 3 X 3 matrix [1,1,2; 1,0,1; 2,1,1].  Gary W. Adamson, Mar 12 2009
a(n) equals term (2,2) in M^n, M = the 3 X 3 matrix [0,1,0; 1,3,1; 0,1,0].  Paul Barry, Sep 18 2009
Starting with "1" = INVERT transform of A002605: (1, 2, 6, 16, 44, ...).
Example: a(3) = 39 = (16, 6, 2, 1) dot (1, 1, 3, 11) = (16 + 6 + 6 + 11). (End)
Pisano period lengths: 1, 1, 4, 1, 24, 4, 48, 2, 12, 24, 30, 4, 12, 48, 24, 4,272, 12, 18, 24, ... .  R. J. Mathar, Aug 10 2012
A007482 is also the number of ways of tiling a 3 X n rectangle with 1 X 1 squares, 2 X 2 squares and 2 X 1 (vertical) dominoes.  R. K. Guy, May 20 2015
With offset 1 (a(0) = 0, a(1) = 1) this is a divisibility sequence.  Michael Somos, Jun 03 2015
Number of elements of size 2^(n) in a fractal generated by the second order reversible cellular automaton, rule 150R (see the reference and the link).  Yuriy Sibirmovsky, Oct 04 2016
a(n) is the number of compositions (ordered partitions) of n into parts 1 (of three kinds) and 2 (of two kinds).  Joerg Arndt, Oct 05 2016
a(n) equals the number of words of length n over {0,1,2,3,4} in which 0 and 1 avoid runs of odd lengths.  Milan Janjic, Jan 08 2017
Start with a single cell at coordinates (0, 0), then iteratively subdivide the grid into 2 X 2 cells and remove the cells that have two '1's in their modulo 3 coordinates. a(n) is the number of cells after n iterations. Cell configuration converges to a fractal with approximate dimension 1.833.  Peter Karpov, Apr 20 2017
This is the Lucas sequence U(P=3,Q=2), and hence for n>=0, a(n+2)/a(n+1) equals the continued fraction 3 + 2/(3 + 2/(3 + 2/(3 + ... + 2/3))) with n 2's.  Greg Dresden, Oct 06 2019


REFERENCES

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Stephen Wolfram, A New Kind of Science, Wolfram Media, 2002, p. 439


LINKS



FORMULA

G.f.: 1/(13*x2*x^2).
a(n) = 3*a(n1) + 2*a(n2).
a(n) = (ap^(n+1)am^(n+1))/(apam), where ap = (3+sqrt(17))/2 and am = (3sqrt(17))/2.
Let b(0) = 1, b(k) = floor(b(k1)) + 2/b(k1); then, for n>0, b(n) = a(n)/a(n1).  Benoit Cloitre, Sep 09 2002
The Hankel transform of this sequence is [1,2,0,0,0,0,0,0,0,...].  Philippe Deléham, Nov 21 2007
a(n) = Sum_{k=0..floor(n/2)} binomial(nk, k)2^k*3^(n2k).  Paul Barry, Apr 23 2005
a(n) =  a(2n) * (2)^(n+1) for all n in Z.  Michael Somos, Jun 03 2015
If c = (3 + sqrt(17))/2, then c^n = (A206776(n) + sqrt(17)*a(n1)) / 2.  Michael Somos, Oct 13 2016
a(n) = 3^n*hypergeom([(1n)/2,n/2], [n], 8/9)) for n>=1.  Peter Luschny, Jun 28 2017
a(n) = round(((sqrt(17) + 3)/2)^(n+1)/sqrt(17)). The distance of the argument from the nearest integer is about 1/2^(n+3).  M. F. Hasler, Jun 16 2019
E.g.f.: (1/17)*exp(3*x/2)*(17*cosh(sqrt(17)*x/2) + 3*sqrt(17)*sinh(sqrt(17)*x/2)).  Stefano Spezia, Oct 07 2019
a(n) = (sqrt(2)*i)^n * ChebyshevU(n, 3*i/(2*sqrt(2))).  G. C. Greubel, Dec 24 2021


EXAMPLE

G.f. = 1 + 3*x + 11*x^2 + 39*x^3 + 139*x^4 + 495*x^5 + 1763*x^6 + ...
For n = 0, (1, ..., 2n) = () is the empty sequence, which is equal to its only subsequence, which satisfies the condition voidly, whence a(0) = 1.
For n = 1, (1, ..., 2n) = (1, 2); among the four subsequences {(), (1), (2), (1,2)} only (1) does not satisfy the condition, whence a(1) = 3.
For n = 2, (1, ..., 2n) = (1, 2, 3, 4); among the sixteen subsequences {(), ..., (1,2,3,4)}, the 5 subsequences (1), (3), (1,3), (2,3,4) and (1,2,3,4) do not satisfy the condition, whence a(2) = 16  5 = 11.
(End)


MAPLE

a := n > `if`(n=0, 1, 3^n*hypergeom([(1n)/2, n/2], [n], 8/9)):


MATHEMATICA

a[n_]:=(MatrixPower[{{1, 4}, {1, 2}}, n].{{1}, {1}})[[2, 1]]; Table[a[n], {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
LinearRecurrence[{3, 2}, {1, 3}, 30] (* Harvey P. Dale, May 25 2013 *)
a[ n_] := Module[ {m = n + 1, s = 1}, If[ m < 0, {m, s} = {m, (2)^m}]; s SeriesCoefficient[ x / (1  3 x  2 x^2), {x, 0, m}]]; (* Michael Somos, Jun 03 2015 *)
a[ n_] := With[{m = n + 1}, If[ m < 0, (2)^m a[ m], Expand[((3 + Sqrt[17])/2)^m  ((3  Sqrt[17])/2)^m ] / Sqrt[17]]]; (* Michael Somos, Oct 13 2016 *)


PROG

(Sage) [lucas_number1(n, 3, 2) for n in range(1, 25)] # Zerinvary Lajos, Apr 22 2009
(PARI) {a(n) = 2*imag(( (3 + quadgen(68)) / 2)^(n+1))}; /* Michael Somos, Jun 03 2015 */
(Haskell)
a007482 n = a007482_list !! (n1)
a007482_list = 1 : 3 : zipWith (+)
(map (* 3) $ tail a007482_list) (map (* 2) a007482_list)
(Maxima) a(n) := if n=0 then 1 elseif n=1 then 3 else 3*a(n1)+2*a(n2);
(Magma) I:=[1, 3]; [n le 2 select I[n] else 3*Self(n1) + 2*Self(n2): n in [1..30]]; // G. C. Greubel, Jan 16 2018


CROSSREFS



KEYWORD

nonn,easy,nice


AUTHOR



STATUS

approved



