

A083329


a(0) = 1; for n > 0, a(n) = 3*2^(n1)  1.


41



1, 2, 5, 11, 23, 47, 95, 191, 383, 767, 1535, 3071, 6143, 12287, 24575, 49151, 98303, 196607, 393215, 786431, 1572863, 3145727, 6291455, 12582911, 25165823, 50331647, 100663295, 201326591, 402653183, 805306367, 1610612735, 3221225471, 6442450943
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OFFSET

0,2


COMMENTS

Apart from leading term (which should really be 3/2), same as A055010.
Binomial transform of A040001. Inverse binomial transform of A053156.
a(n) = A105728(n+1,2).  Reinhard Zumkeller, Apr 18 2005
Row sums of triangle A133567.  Gary W. Adamson, Sep 16 2007
Row sums of triangle A135226.  Gary W. Adamson, Nov 23 2007
a(n) = number of partitions Pi of [n+1] (in standard increasing form) such that the permutation Flatten[Pi] avoids the patterns 213 and 312. Example: a(3)=11 counts all 15 partitions of [4] except 13/24, 13/2/4 which contain a 213 and 14/23, 14/2/3 which contain a 312. Here "standard increasing form" means the entries are increasing in each block and the blocks are arranged in increasing order of their first entries.  David Callan, Jul 22 2008
An elephant sequence, see A175654. For the corner squares four A[5] vectors, with decimal values 42, 138, 162, 168, lead to this sequence. For the central square these vectors lead to the companion sequence A003945.  Johannes W. Meijer, Aug 15 2010
The binary representation of a(n) has n+1 digits, where all digits are 1's except digit n1. For example: a(4) = 23 = 10111 (2).  Omar E. Pol, Dec 02 2012
Row sums of triangle A209561.  Reinhard Zumkeller, Dec 26 2012
If a Stern's sequence based enumeration system of positive irreducible fractions is considered (for example, A007305/A047679, A162909/A162910, A071766/A229742, A245325/A245326, ...), and if it is organised by blocks or levels (n) with 2^n terms (n>=0), and the fractions, term by term, are summed at each level n, then the resulting sequence of integers is a(n)+1/2, apart from leading term (which should be 1/2).  Yosu Yurramendi, May 23 2015


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Sergey Kitaev, Jeffrey Remmel, Mark Tiefenbruck, Quadrant Marked Mesh Patterns in 132Avoiding Permutations II, Electronic Journal of Combinatorial Number Theory, Volume 15 #A16. (arXiv:1302.2274)
Index entries for linear recurrences with constant coefficients, signature (3,2).


FORMULA

a(n) = (3*2^n  2 + 0^n)/2.
G.f.: (1x+x^2)/((1x)*(12*x)).
E.g.f.: (3*exp(2*x)2*exp(x)+exp(0))/2.
a(0) = 1, a(n) = sum of all previous terms + n.  Amarnath Murthy, Jun 20 2004
a(n) = 3*a(n1)2*a(n2) for n>2, a(0)=1, a(1)=2, a(2)=5.  Philippe Deléham, Nov 29 2013
From Bob Selcoe, Apr 25 2014: (Start)
a(n) = (...((((((1)+1)*2+1)*2+1)*2+1)*2+1)...), with n+1 1's, n >= 0.
a(n) = 2*a(n1) + 1, n >= 2.
a(n) = 2^n + 2^(n1)  1, n >= 2. (End)


EXAMPLE

a(0) = (3*2^0  2 + 0^0)/2 = 2/2 = 1 (use 0^0=1).


MAPLE

seq(ceil((2^i+2^(i+1)2)/2), i=0..31); # Zerinvary Lajos, Oct 02 2007


MATHEMATICA

a[1] = 2; a[n_] := 2a[n  1] + 1; Table[ a[n], {n, 31}] (* Robert G. Wilson v, May 04 2004 *)
Join[{1}, LinearRecurrence[{3, 2}, {2, 5}, 40]] (* Vincenzo Librandi, Jan 01 2016 *)


PROG

(Haskell)
a083329 n = a083329_list !! n
a083329_list = 1 : iterate ((+ 1) . (* 2)) 2
 Reinhard Zumkeller, Dec 26 2012, Feb 22 2012
(PARI) a(n)=(3*2^n2+0^n)/2 \\ Charles R Greathouse IV, Sep 24 2015
(MAGMA) [1] cat [3*2^(n1)1: n in [1..40]]; // Vincenzo Librandi, Jan 01 2016


CROSSREFS

Essentially the same as A055010 and A052940.
Cf. A000225, A052955, A133567, A135226.
Cf. A007505 (primes).
Cf. A266550 (independence number of the nMycielski graph).
Sequence in context: A086219 A055010 A153893 * A266550 A081973 A055496
Adjacent sequences: A083326 A083327 A083328 * A083330 A083331 A083332


KEYWORD

easy,nonn


AUTHOR

Paul Barry, Apr 27 2003


EXTENSIONS

The generating function corrected by Martin Griffiths (griffm(AT)essex.ac.uk), Dec 01 2009


STATUS

approved



