
COMMENTS

Partial sums of A000244. Values of base 3 strings of 1's.
a(n) = (3^n1)/2 is also the number of different nonparallel lines determined by pair of vertices in the n dimensional hypercube. Example: when n = 2 the square has 4 vertices and then the relevant lines are: x = 0, y = 0, x = 1, y = 1, y = x, y = 1x and when we identify parallel lines only 4 remain: x = 0, y = 0, y = x, y = 1x so a(2) = 4.  Noam Katz (noamkj(AT)hotmail.com), Feb 11 2001
Also number of 3block bicoverings of an nset (if offset is 1, cf. A059443).  Vladeta Jovovic, Feb 14 2001
3^a(n) is the highest power of 3 dividing (3^n)!.  Benoit Cloitre, Feb 04 2002
Apart from a(0) term, maximum number of coins among which a lighter or heavier counterfeit coin can be detected by n weighings.  Tom Verhoeff (Tom.Verhoeff(AT)acm.org), Jun 22 2002
n such that A001764(n) is not divisible by 3.  Benoit Cloitre, Jan 14 2003
Consider the mapping f(a/b) = (a + 2b)/(2a + b). Taking a = 1, b = 2 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the sequence 1/2, 4/5, 13/14, 40/41, ... converging to 1. Sequence contains the numerators = (3^n1)/2. The same mapping for N, i.e., f(a/b) = (a + Nb)/(a+b) gives fractions converging to N^(1/2).  Amarnath Murthy, Mar 22 2003
Binomial transform of A000079 (with leading zero).  Paul Barry, Apr 11 2003
With leading zero, inverse binomial transform of A006095.  Paul Barry, Aug 19 2003
Number of walks of length 2*n+2 in the path graph P_5 from one end to the other one. Example: a(2)=4 because in the path ABCDE we have ABABCDE, ABCBCDE, ABCDCDE and ABCDEDE.  Emeric Deutsch, Apr 02 2004
The number of triangles of all sizes (not counting holes) in Sierpiński's triangle after n inscriptions.  Lee Reeves (leereeves(AT)fastmail.fm), May 10 2004
Number of (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < 6 and s(i)  s(i1) = 1 for i = 1,2,....,2*n+1, s(0) = 1, s(2n+1) = 4.  Herbert Kociemba, Jun 10 2004
Number of nondegenerate rightangled incongruent integeredged Heron triangles whose circumdiameter is the product of n distinct primes of shape 4k + 1.  Alex Fink and R. K. Guy, Aug 18 2005
Also numerator of the sum of the reciprocals of the first n powers of 3, with A000244 being the sequence of denominators. With the exception of n < 2, the base 10 digital root of a(n) is always 4. In base 3 the digital root of a(n) is the same as the digital root of n.  Alonso del Arte, Jan 24 2006
The sequence 3*a(n), n>=1, gives the number of edges of the Hanoi graph H_3^{n} with 3 pegs and n>=1 discs.  Daniele Parisse (daniele.parisse(AT)eads.com), Jul 28 2006
Numbers n such that a(n) is prime are listed in A028491 = {3,7,13,71,103,541,1091,...}. 2^(m+1) divides a(2^m*k) for m>0. 5 divides a(4k). 5^2 divides a(20k). 7 divides a(6k). 7^2 divides a(42k). 11^2 divides a(5k). 13 divides a(3k). 17 divides a(16k). 19 divides a(18k). 1093 divides a(7k). 41 divides a(8k). p divides a((p1)/5) for prime p = {41,431,491,661,761,1021,1051,1091,1171,...}. p divides a((p1)/4) for prime p = {13,109,181,193,229,277,313,421,433,541,...}. p divides a((p1)/3) for prime p = {61,67,73,103,151,193,271,307,367,...} = A014753, 3 and 3 are both cubes (one implies other) mod these primes p=1 mod 6. p divides a((p1)/2) for prime p = {11,13,23,37,47,59,61,71,73,83,97,...} = A097933(n). p divides a(p1) for prime p>7. p^2 divides a(p*(p1)k) for all prime p except p = 3. p^3 divides a(p*(p1)*(p2)k) for prime p = 11.  Alexander Adamchuk, Jan 22 2007
Let P(A) be the power set of an nelement set A. Then a(n) = the number of [unordered] pairs of elements {x,y} of P(A) for which x and y are disjoint [and both nonempty]. Wieder calls these "disjoint usual 2combinations".  Ross La Haye, Jan 10 2008 [This is because each of the elements of {1,2,...,n} can be in the first subset, in the second or in neither. Because there are three options for each, the total number of options is 3^n. However, since the sets being empty is not an option we subtract 1 and since the subsets are unordered we then divide by 2! (The number of ways two objects can be arranged.) Thus we obtain (3^n1)/2=a(n).  Chayim Lowen, Mar 03 2015]
Starting with offset 1 = binomial transform of A003945: (1, 3, 6, 12, 24, ...) and double bt of (1, 2, 1, 2, 1, 2, ...); equals polcoeff inverse of (1, 4, 3, 0, 0, 0, ...).  Gary W. Adamson, May 28 2009
Also the constant of the polynomials C(x)=3x+1 that form a sequence by performing this operation repeatedly and taking the result at each step as the input at the next.  Nishant Shukla (n.shukla722(AT)gmail.com), Jul 11 2009
It appears that this is A120444(3^n1) = A004125(3^n)  A004125(3^n1), where A004125 is the sum of remainders of n mod k for k=1,2,3,...,n.  John W. Layman, Jul 29 2009
Subsequence of A134025; A171960(a(n)) = a(n).  Reinhard Zumkeller, Jan 20 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=3, (i>1), A[i,i1]=1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A).  Milan Janjic, Jan 27 2010
This is the sequence A(0,1;2,3;2) = A(0,1;4,3;0) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below.  Wolfdieter Lang, Oct 18 2010
It appears that if s(n) is a first order rational sequence of the form s(0)=0, s(n)= (2*s(n1)+1)/(s(n1)+2), n>0, then s(n)= a(n)/(a(n)+1).  Gary Detlefs, Nov 16 2010
This sequence also describes the total number of moves to solve the [RED ; BLUE ; BLUE] or [RED ; RED ; BLUE] precolored Magnetic Towers of Hanoi puzzle (cf. A183111  A183125).
From Adi Dani, Jun 08 2011: (Start)
a(n) is number of compositions of odd numbers into n parts <3. For example, a(3)=13 and there are 13 compositions odd numbers into 3 parts <3:
1: (0,0,1),(0,1,0),(1,0,0);
3: (0,1,2),(0,2,1),(1,0,2),(1,2,0),(2,0,1),(2,1,0),(1,1,1);
5: (1,2,2),(2,1,2),(2,2,1).
(End)
For n > 1: A008344(a(n)) = 0.  Reinhard Zumkeller, May 09 2012
Pisano period lengths: 1, 2, 1, 2, 4, 2, 6, 4, 1, 4, 5, 2, 3, 6, 4, 8, 16, 2, 18, 4, ... .  R. J. Mathar, Aug 10 2012
A085059(a(n)) = 1 for n > 0.  Reinhard Zumkeller, Jan 31 2013
a(n) is the total number of holes (triangles removed) after the nth step of a Sierpiński triangle production.  Ivan N. Ianakiev, Oct 29 2013
a(n) solves Sum_{j=a(n)+1..a(n+1)} j = k^2 for some integer k, given a(0) = 0 and requiring smallest a(n+1) > a(n). Corresponding k = 3^n.  Richard R. Forberg, Mar 11 2015
