A083329 a(0) = 1; for n > 0, a(n) = 3*2^(n-1) - 1.

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

Offset: 0

Author

Paul Barry, Apr 27 2003

Keywords

• easy
• nonn
• changed

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 2-1-3 and 3-1-2. Example: a(3)=11 counts all 15 partitions of [4] except 13/24, 13/2/4 which contain a 2-1-3 and 14/23, 14/2/3 which contain a 3-1-2. 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 n-1. 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 organized 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

For n >= 2, A083329(n) in binary representation is a string [101..1], also 10 followed with (n-1) 1's. For n >= 3, A036563(n) in binary representation is a string [1..101], also (n-2) 1's followed with 01. Thus A083329(n) is a reflection of the binary representation of A036563(n+1). Example: A083329(5) = 101111 in binary, A036563(6) = 111101 in binary. - Ctibor O. Zizka, Nov 06 2018

For n > 0, a(n) is the minimum number of turns in (n+1)-dimensional Euclidean space needed to visit all 2^(n+1) vertices of the (n+1)-cube (e.g., {0,1}^(n+1)) and return to the starting point, moving along straight-line segments between turns (turns may occur elsewhere in R^(n+1)). - Marco Ripà, Aug 14 2025

Examples

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

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• M. Griffiths, I. Mezo, A generalization of Stirling Numbers of the Second Kind via a special multiset, JIS 13 (2010) #10.2.5.
• Sergey Kitaev, Jeffrey Remmel, Mark Tiefenbruck, Quadrant Marked Mesh Patterns in 132-Avoiding Permutations II, Electronic Journal of Combinatorial Number Theory, Volume 15 #A16. (arXiv:1302.2274)
• Agustín Moreno Cañadas, Hernán Giraldo, Gabriel Bravo Rios, On the Number of Sections in the Auslander-Reiten Quiver of Algebras of Dynkin Type, Far East Journal of Mathematical Sciences (FJMS), Vol. 101, No. 8 (2017), pp. 1631-1654.
• Roberto Rinaldi and Marco Ripà, Optimal cycles enclosing all the nodes of a k-dimensional hypercube, arXiv:2212.11216 [math.CO], 2022.
• Index entries for linear recurrences with constant coefficients, signature (3,-2).

Crossrefs

Essentially the same as A055010 and A052940.

Cf. A000225, A052955, A133567, A135226.

Cf. A007505 (primes).

Cf. A266550 (independence number of the n-Mycielski graph).

Programs

Haskell

a083329 n = a083329_list !! n
a083329_list = 1 : iterate ((+ 1) . (* 2)) 2
-- Reinhard Zumkeller, Dec 26 2012, Feb 22 2012

Magma

[1] cat [3*2^(n-1)-1: n in [1..40]]; // Vincenzo Librandi, Jan 01 2016

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 *)

PARI

a(n)=(3*2^n-2+0^n)/2 \\ Charles R Greathouse IV, Sep 24 2015

Formula

a(n) = (3*2^n - 2 + 0^n)/2.

G.f.: (1-x+x^2)/((1-x)*(1-2*x)). [corrected by Martin Griffiths, Dec 01 2009]

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

a(n) = 2^n + 2^(n-1) - 1, n >= 2. (End)

a(n) = A086893(n) + A061547(n+1), n > 0. - Yosu Yurramendi, Jan 16 2017