

A000244


Powers of 3: a(n) = 3^n.
(Formerly M2807 N1129)


820



1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467, 3486784401, 10460353203, 31381059609, 94143178827, 282429536481, 847288609443, 2541865828329, 7625597484987
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OFFSET

0,2


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 such 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
With p(n) being the number of integer partitions of n, p(i) being the number of parts of the ith partition of n, d(i) being the number of different parts of the ith partition of n, m(i, j) being the multiplicity of the jth part of the ith partition of n, Sum_{i = 1..p(n)} being the sum over i and Product_{j = 1..d(i)} being the 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
With regard to the comment by Ross La Haye:
Cf. A001047 if either nonempty subsets are considered or x is a proper subset of y.
Cf. a(n+1) in A028243 if nonempty subsets are considered and x is a proper subset of y. (End)
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
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
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
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 ((p  1) + 1)^(n  1) = Sum_{k = 0..n  1} binomial(n  1, k) (p  1)^k 1^(n  1  k) = Sum_{k = 1..n} binomial(n  1, k  1) (p  1)^(k  1).
(End)
Also, first and least element of the matrix [1, sqrt(2); sqrt(2), 2]^(n+1).  M. F. Hasler, Nov 25 2011
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..k  1} m(n, c) + Sum_{r = 0..n  1} 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..k  1} m(n, c) + Sum_{r = 0..n  1} 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
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 3term 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 there are no further terms.  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 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
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
Also the number of independent vertex sets and vertex covers in the nladder rung graph n P_2.  Eric W. Weisstein, Sep 21 2017
Also the number of (not necessarily maximal) cliques in the ncocktail party graph.  Eric W. Weisstein, Nov 29 2017
a(n1) is the number of 2compositions of n; see Hopkins & Ouvry reference.  Brian Hopkins, Aug 15 2020
a(n) is the number of faces of any dimension (vertices, edges, square faces, etc.) of the ndimensional hypercube. For example, the 0dimensional hypercube is a point, and its only face is itself. The 1dimensional hypercube is a line, which has two vertices and an edge. The 2dimensional hypercube is a square, which has four vertices, four edges, and a square face.  Kevin Long, Mar 14 2023
Number of pairs (A,B) of subsets of M={1,2,...,n} with union(A,B)=M. For nonempty subsets cf. A058481.  Manfred Boergens, Mar 28 2023
a(n) is the number of disjunctive clauses of n variables up to equivalence. A disjunctive clause is a propositional formula of the form l_1 OR ... OR l_m, where l_1, ..., l_m are distinct elements in {x_1, ..., x_n, NOT x_1, ..., NOT x_n} for n variables x_1, ... x_n, and no x_i and NOT x_i appear at the same time. For each 1 <= i <= n, we can have neither of x_i or NOT x_i, only x_i or only NOT x_i appearing in a disjunctive clause, so the number of such clauses is 3^n. Viewing the propositional formulas of n variables as functions {0,1}^n > {0,1}, a disjunctive clause corresponds to a function f such that the inverse image of 0 is of the form A_1 X ... X A_n, where A_i is nonempty for all 1 <= i <= n. Since each A_i has 3 choices ({0}, {1} or {0,1}), we also find that the number of disjunctive clauses of n variables is 3^n.
Equivalently, a(n) is the number of conjunctive clauses of n variables. (End)
The finite subsequence a(2), a(3), a(4), a(5) = 9, 27, 81, 243 is one of only two geometric sequences that can be formed with all interior angles (all integer, in degrees) of a simple polygon. The other sequence is a subsequence of A007283 (see comment there).  Felix Huber, Feb 15 2024


REFERENCES

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Doron Zeilberger, The Amazing 3^n Theorem and its even more Amazing Proof [Discovered by Xavier G. Viennot and his École Bordelaise gang], arXiv:1208.2258, 2012.


LINKS

Eric Weisstein's World of Mathematics, Clique


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) = 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
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
2/3 + 3/3^2 + 2/3^3 + 3/3^4 + 2/3^5 + ... = 9/8. [Jolley, Summation of Series, Dover, 1961]
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(n1) = binomial(2*n1, n) + Sum_{k >= 1} binomial(2*n, n+3*k)*(1)^k.  Greg Dresden, Oct 14 2022
G.f.: Sum_{k >= 0} x^k/(12*x)^(k+1).  Kevin Long, Mar 14 2023


EXAMPLE

G.f. = 1 + 3*x + 9*x^2 + 27*x^3 + 81*x^4 + 243*x^5 + 729*x^6 + 2187*x^7 + ...


MAPLE

A000244 := n>3^n; [ seq(3^n, n=0..50) ];


MATHEMATICA

CoefficientList[Series[1/(1  3 x), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)


PROG

(Haskell)
(Maxima) makelist(3^n, n, 0, 30); /* Martin Ettl, Nov 05 2012 */
(Scala) val powersOf3: LazyList[BigInt] = LazyList.iterate(1: BigInt)(_ * 3)
(Python)


CROSSREFS

Cf. A008776 (2*a(n), and first differences).
The following are parallel families: A000079 (2^n), A004094 (2^n reversed), A028909 (2^n sorted up), A028910 (2^n sorted down), A036447 (double and reverse), A057615 (double and sort up), A263451 (double and sort down); A000244 (3^n), A004167 (3^n reversed), A321540 (3^n sorted up), A321539 (3^n sorted down), A163632 (triple and reverse), A321542 (triple and sort up), A321541 (triple and sort down).


KEYWORD

nonn,nice,easy,core


AUTHOR



STATUS

approved



