
COMMENTS

Same as Pisot sequences E(1,3), L(1,3), P(1,3), T(1,3). Essentially same as Pisot sequences E(3,9), L(3,9), P(3,9), T(3,9). See A008776 for definitions of Pisot sequences.
Number of (s(0), s(1), ..., s(2n+2)) such that 0 < s(i) < 6 and s(i)  s(i1) = 1 for i = 1, 2, ..., 2n + 2, s(0) = 1, s(2n+2) = 3.  Herbert Kociemba, Jun 10 2004
a(1) = 1, a(n+1) is the least number so that there are a(n) even numbers between a(n) and a(n+1). Generalization for the sequence of powers of k: 1, k, k^2, k^3, k^4, ... There are a(n) multiples of k1 between a(n) and a(n+1).  Amarnath Murthy, Nov 28 2004
a(n) = sum of (n+1)th row in Triangle A105728.  Reinhard Zumkeller, Apr 18 2005
With p(n) = the number of integer partitions of n, p(i) = the number of parts of the ith partition of n, d(i) = the number of different parts of the ith partition of n, m(i,j) = multiplicity of the jth part of the ith partition of n, Sum_{i=1..p(n)} = sum over i and Product_{j=1..d(i)} = product over j one has: a(n) = Sum_{i=1..p(n)} (p(i)!/(Product_{j=1..d(i)} m(i, j)!))*2^(p(i)  1).  Thomas Wieder, May 18 2005
For any k > 1 in the sequence, k is the first prime power appearing in the prime decomposition of repunit R_k, i.e., of A002275(k).  Lekraj Beedassy, Apr 24 2006
a(n1) is the number of compositions of compositions. In general, (k+1)^(n1) is the number of klevels nested compositions (e.g., 4^(n1) is the number of compositions of compositions of compositions, etc.). Each of the n  1 spaces between elements can be a break for one of the k levels, or not a break at all.  Franklin T. AdamsWatters, Dec 06 2006
Let S be a binary relation on the power set P(A) of a set A having n = A elements such that for every element x, y of P(A), xSy if x is a subset of y. Then a(n) = S.  Ross La Haye, Dec 22 2006
If X_1, X_2, ..., X_n is a partition of the set {1, 2, ..., 2*n} into blocks of size 2 then, for n >= 1, a(n) is equal to the number of functions f : {1, 2, ..., 2*n} > {1, 2} such that for fixed y_1, y_2, ..., y_n in {1, 2} we have f(X_i) <> {y_i}, (i = 1, 2, ..., n).  Milan Janjic, May 24 2007
This is a general comment on all sequences of the form a(n) = [(2^k)1]^n for all positive integers k. Example 1.1.16 of Stanley's "Enumerative Combinatorics" offers a slightly different version. a(n) in the number of functions f:[n] into P([k])  {}. a(n) is also the number of functions f:[k] into P([n]) such that the generalized intersection of f(i) for all i in [k] is the empty set. Where [n] = {1, 2, ..., n}, P([n]) is the power set of [n] and {} is the empty set.  Geoffrey Critzer, Feb 28 2009
a(n) = A064614(A000079(n)) and A064614(m)<a(n) for m < A000079(n).  Reinhard Zumkeller, Feb 08 2010
3^(n+1) = (1, 2, 2, 2, ...) dot (1, 1, 3, 9, ..., 3^n); e.g., 3^3 = 27 = (1, 2, 2, 2) dot (1, 1, 3, 9) = (1 + 2 + 6 + 18).  Gary W. Adamson, May 17 2010
a(n) is the number of generalized compositions of n when there are 3*2^i different types of i, (i = 1, 2, ...).  Milan Janjic, Sep 24 2010
For n >= 1, a(n1) is the number of generalized compositions of n when there are 2^(i1) different types of i, (i = 1, 2, ...).  Milan Janjic, Sep 24 2010
The sequence in question ("Powers of 3") also describes the number of moves of the kth disk solving the [RED ; BLUE ; BLUE] or [RED ; RED ; BLUE] precolored Magnetic Tower of Hanoi puzzle (cf. A183111  A183125).
a(n) is the number of Stern polynomials of degree n. See A057526.  T. D. Noe, Mar 01 2011
Positions of records in the number of odd prime factors, A087436.  JuriStepan Gerasimov, Mar 17 2011
Sum of coefficients of the expansion of (1+x+x^2)^n.  Adi Dani, Jun 21 2011
a(n) is the number of compositions of n elements among {0, 1, 2}; e.g., a(2) = 9 since there are the 9 compositions 0 + 0, 0 + 1, 1 + 0, 0 + 2, 1 + 1, 2 + 0, 1 + 2, 2 + 1, and 2 + 2. [From Adi Dani, Jun 21 2011; modified by editors.]
Except the first two terms, these are odd numbers n such that no x with 2 <= x <= n  2 satisfy x^(n1) == 1 (mod n).  Arkadiusz Wesolowski, Jul 03 2011
The compositions of n in which each natural number is colored by one of p different colors are called pcolored compositions of n. For n >= 1, a(n) equals the number of 3colored compositions of n such that no adjacent parts have the same color.  Milan Janjic, Nov 17 2011
Explanation from David Applegate, Feb 20 2017. (Start) Since the preceding comment appears in a large number of sequences, it might be worth adding a proof.
The number of compositions of n into exactly k parts is binomial(n1,k1).
For a pcolored composition of n such that no adjacent parts have the same color, there are exactly p choices for the color of the first part, and p1 choices for the color of each additional part (any color other than the color of the previous one). So, for a partition into k parts, there are p (p1)^(k1) valid colorings.
Thus the number of pcolored compositions of n into exactly k parts such that no adjacent parts have the same color is binomial(n1,k1) p (p1)^(k1).
The total number of pcolored compositions of n such that no adjacent parts have the same color is then
sum_{k=1..n} binomial(n1,k1) * p * (p1)^(k1) = p^n.
To see this, note that the binomial expansion of ((p1)+1)^(n1) = sum_{k=0..n1} binomial(n1,k) (p1)^k 1^(n1k) = sum_{k=1..n} binomial(n1,k1) (p1)^(k1).
(End)
Also, first and least element of the matrix [1, sqrt(2); sqrt(2), 2]^(n+1).  M. F. Hasler, Nov 25 2011
Onehalf of the row sums of the triangular version of A035002.  J. M. Bergot, Jun 10 2013
Form an array with m(0,n) = m(n,0) = 2^n; m(i,j) equals the sum of the terms to the left of m(i,j) and the sum of the terms above m(i,j), which is m(i,j) = Sum_{k=0..j1} m(i,k) + Sum_{k=0..i1} m(k,j). The sum of the terms in antidiagonal(n+1) = 4*a(n).  J. M. Bergot, Jul 10 2013
a(n) = A007051(n+1)  A007051(n), and A007051 are the antidiagonal sums of an array defined by m(0,k) = 1 and m(n,k) = Sum_{c=0..k1} m(n,c) + Sum_{r=0..n1} m(r,k), which is the sum of the terms to left of m(n,k) plus those above m(n,k). m(1,k) = A000079(k); m(2,k) = A045623(k+1); m(k+1,k) = A084771(k).  J. M. Bergot, Jul 16 2013
Define an array to have m(0,k) = 2^k and m(n,k) = Sum_{c=0..k1} m(n,c) + Sum_{r=0..n1} m(r,k), which is the sum of the terms to the left of m(n,k) plus those above m(n,k). Row n = 0 of the array comprises A000079, column k = 0 comprises A011782, row n = 1 comprises A001792. Antidiagonal sums of the array are a(n): 1 = 3^0, 1 + 2 = 3^1, 2 + 3 + 4 = 3^2, 4 + 7 + 8 + 8 = 3^3.  J. M. Bergot, Aug 02 2013
The sequence with interspersed zeros and o.g.f. x/(1 3*x^2), A(2*k) = 0, A(2*k+1) = 3^k = a(k), k >= 0, can be called hexagon numbers. This is because the algebraic number rho(6) = 2*cos(Pi/6) = sqrt(3) of degree 2, with minimal polynomial C(6, x) = x^2  3 (see A187360, n = 6), is the length ratio of the smaller diagonal and the side in the hexagon. Hence rho(6)^n = A(n1)*1 + A(n)*rho(6), in the power basis of the quadratic number field Q(rho(6)). One needs also A(1) = 1. See also a Dec 02 2010 comment and the P. Steinbach reference given in A049310.  Wolfdieter Lang, Oct 02 2013
Numbers n such that sigma(3n) = 3n + sigma(n).  Jahangeer Kholdi, Nov 23 2013
All powers of 3 are perfect totient numbers (A082897), since phi(3^n) = 2 * 3^(n  1) for n > 0, and thus Sum_{i=0..n} phi(3^i) = 3^n.  Alonso del Arte, Apr 20 2014
The least number k > 0 such that 3^k ends in n consecutive decreasing digits is a 3number sequence given by {1, 13, 93}. The consecutive increasing digits are {3, 23, 123}. There are 100 different 3digit endings for 3^k. There are no kvalues such that 3^k ends in '012', '234', '345', '456', '567', '678', or '789'. The kvalues for which 3^k ends in '123' are given by 93 mod 100. For k = 93 + 100*x, the digit immediately before the run of '123' is {9, 5, 1, 7, 3, 9, 5, 1, 3, 7, ...} for x = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...}, respectively. Thus, we see the digit before '123' will never be a 0. So, this sequence is finite and full.  Derek Orr, Jul 03 2014
All elements of A^n where A=(1,1,1;1,1,1;1,1,1).  David Neil McGrath, Jul 23 2014
Counts all walks of length n (open or closed) on the vertices of a triangle containing a loop at each vertex starting from any given vertex.  David Neil McGrath, Oct 03 2014
a(n) counts walks (closed) on the graph G(1vertex;1loop,1loop,1loop).  David Neil McGrath, Dec 11 2014
2*a(n2) counts all permutations of a solitary closed walk of length (n) from the vertex of a triangle that contains 2 loops on each of the remaining vertices. In addition, C(m,k)=2*(2^m)*B(m+k2,m) counts permutations of walks that contain (m) loops and (k) arcs.  David Neil McGrath, Dec 11 2014
a(n) is the number of all elements (mdimensional faces) in an ndimensional cube or in an ndimensional orthoplex (or crosspolytope) (m>=0, m<=n).  Sergey Pavlov, 15 Aug 2015
a(n) is the sum of the coefficients of the nth layer of Pascal's pyramid (a.k.a., Pascal's tetrahedron  see A046816).  Bob Selcoe, Apr 02 2016
Numbers n such that the trinomial x^(2*n) + x^n + 1 is irreducible over GF(2). Of these only the trinomial for n=1 is primitive.  Joerg Arndt, May 16 2016
Let n = 44a + b, where a = floor(n/44), b = n mod 44, b is even, b < 44. Then, for any even n such that 0 < n < 306, a(n) has u digits where u = n/2  a, or u = n/2  floor(n/44). While b = 0, or n == 0 (mod 44), it seems that, at least for n < 39645 (n is even), u = 21a  floor(a/k), or u = 21n/44  floor(n/(44k)), where k = 150 + 120/a.  Sergey Pavlov, Feb 02 2017
Satisfies Benford's law [BergerHill, 2011].  N. J. A. Sloane, Feb 08 2017
a(n1) is also the number of compositions of n if the parts can be runs of any length from 1 to n, and can contain any integers from 1 to n.  Gregory L. Simay, May 26 2017


FORMULA

a(n) = 3^n.
a(0) = 1; a(n) = 3*a(n1).
G.f.: 1/(13*x).
E.g.f.: exp(3*x).
a(n) = n!*Sum_{i + j + k = n, i, j, k >= 0} 1/(i!*j!*k!).  Benoit Cloitre, Nov 01 2002
a(n) = Sum_{k = 0..n} 2^k*binomial(n, k), binomial transform of A000079.
a(n) = A090888(n, 2).  Ross La Haye, Sep 21 2004
a(n) = 2^(2n)  A005061(n).  Ross La Haye, Sep 10 2005
a(n) = A112626(n, 0).  Ross La Haye, Jan 11 2006
Hankel transform of A007854 = [1, 3, 12, 51, 222, 978, 4338, ...].  Philippe Deléham, Nov 26 2006
a(n) = 2*StirlingS2(n+1,3) + StirlingS2(n+2,2) = 2*(StirlingS2(n+1,3) + StirlingS2(n+1,2)) + 1.  Ross La Haye, Jun 26 2008
a(n) = 2*StirlingS2(n+1, 3) + StirlingS2(n+2, 2) = 2*(StirlingS2(n+1, 3) + StirlingS2(n+1, 2)) + 1.  Ross La Haye, Jun 09 2008
Sum_{n >= 0} 1/a(n) = 3/2.  Gary W. Adamson, Aug 29 2008
If p[i] = fibonacci(2i2) and if A is the Hessenberg matrix of order n defined by: A[i, j] = p[ji+1], (i <= j), A[i, j] = 1, (i = j+1), and A[i, j] = 0 otherwise. Then, for n >= 1, a(n1) = det A.  Milan Janjic, May 08 2010
G.f. A(x) = M(x)/(1M(x))^2, M(x)  o.g.f for Motzkin numbers (A001006).  Vladimir Kruchinin, Aug 18 2010
a(n) = A133494(n+1).  Arkadiusz Wesolowski, Jul 27 2011
2/3 + 3/3^2 + 2/3^3 + 3/3^4 + 2/3^5 + ... = 9/8. [Jolley, Summation of Series, Dover, 1961]
a(n) = Sum_{k=0..n} A207543(n,k)*4^(nk).  Philippe Deléham, Feb 25 2012
a(n) = Sum_{k=0..n} A125185(n,k).  Philippe Deléham, Feb 26 2012
Sum_{n > 0} mobius(n)/a(n) = 0.181995386702633887827... (see A238271).  Alonso del Arte, Aug 09 2012. See also the sodium 3s orbital energy in table V of J. Chem. Phys. 53 (1970) 348.
a(0) = 1, a(1) = 3; for n > 1, a(n) = (k+3)*a(n1)  3*k*a(n2) for all k.  Vincenzo Librandi, Feb 16 2014
a(n) = 1/3 + Sum_{i=0..n} a(i)  a(i1).  Wesley Ivan Hurt, Jul 04 2014
a(n) = 2*a(n1) + 3*a(n2).  Tim Fulford, Sep 27 2016
