

A003462


a(n) = (3^n  1)/2.
(Formerly M3463)


281



0, 1, 4, 13, 40, 121, 364, 1093, 3280, 9841, 29524, 88573, 265720, 797161, 2391484, 7174453, 21523360, 64570081, 193710244, 581130733, 1743392200, 5230176601, 15690529804, 47071589413, 141214768240, 423644304721, 1270932914164
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OFFSET

0,3


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 = 1  x so a(2) = 4.  Noam Katz (noamkj(AT)hotmail.com), Feb 11 2001
3^a(n) is the highest power of 3 dividing (3^n)!.  Benoit Cloitre, Feb 04 2002
Apart from the a(0) and a(1) terms, maximum number of coins among which a lighter or heavier counterfeit coin can be identified (but not necessarily labeled as heavier or lighter) by n weighings.  Tom Verhoeff, Jun 22 2002, updated Mar 23 2017
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
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, 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]
Also, still with P(A) being the power set of a nelement set A, a(n) is the number of 2element subsets {x,y} of P(A) such that the union of x and y is equal to A. Cf. A341590.  Fabio Visonà, Feb 20 2021
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
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 Gary Detlefs, and treated as A(a, b; c, d; k) in the Wolfdieter 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).
a(n) is number of compositions of odd numbers into n parts less than 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)
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
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
a(n+1) equals the number of words of length n over {0, 1, 2, 3} avoiding 01, 02 and 03.  Milan Janjic, Dec 17 2015
For n >= 1, a(n) is also the total number of words of length n, over an alphabet of three letters, such that one of the letters appears an odd number of times (See A006516 for 4 letter words, and the Balakrishnan reference there).  Wolfdieter Lang, Jul 16 2017
Also, the number of maximal cliques, maximum cliques, and cliques of size 4 in the nApollonian network.  Andrew Howroyd, Sep 02 2017
For n > 1, the number of triangles (cliques of size 3) in the (n1)Apollonian network.  Andrew Howroyd, Sep 02 2017
a(n) is the largest number that can be represented with n trits in balanced ternary. Correspondingly, a(n) is the smallest number that can be represented with n trits in balanced ternary.  Thomas König, Apr 26 2020
These form Sierpinski nestingstars, which alternate pattern on 3^n+1/2 star numbers A003154, based on the square configurations of 9^n. The partial sums of 3^n are delineated according to the geometry of a hexagram, see illustrations in links. (3*a(n1) + 1) create Sierpinskiantitriangles, representing the number of holes in a (n+1) Sierpinski triangle (see illustrations).  John Elias, Oct 18 2021
For n > 1, a(n) is the number of iterations necessary to calculate the hyperbolic functions with CORDIC.  Mathias Zechmeister, Jul 26 2022
For all n >= 0, Sum_{k=a(n)+1..a(n+1)} 1/k < Sum_{j=a(n+1)+1..a(n+2)} 1/j. These are the minimal points which partition the infinite harmonic series into a monotonically increasing sequence. Each partition approximates log(3) from below as n tends to infinity.  Joseph Wheat, Apr 15 2023


REFERENCES

J. G. Mauldon, Strong solutions for the counterfeit coin problem, IBM Research Report RC 7476 (#31437) 9/15/78, IBM Thomas J. Watson Research Center, P. O. Box 218, Yorktown Heights, N. Y. 10598.
Paulo Ribenboim, The Book of Prime Number Records, SpringerVerlag, NY, 2nd ed., 1989, p. 60.
Paulo Ribenboim, The Little Book of Big Primes, SpringerVerlag, NY, 1991, p. 53.
Amir Sapir, The Tower of Hanoi with Forbidden Moves, The Computer J. 47 (1) (2004) 20, case threeina row, sequence a(n).
Robert Sedgewick, Algorithms, 1992, pp. 109.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

Max A. Alekseyev and Toby Berger, Solving the Tower of Hanoi with Random Moves. In: J. Beineke, J. Rosenhouse (eds.) The Mathematics of Various Entertaining Subjects: Research in Recreational Math, Princeton University Press, 2016, pp. 6579. ISBN 9780691164038
G. Kreweras, Sur les éventails de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #15 (1970), 341. [Annotated scanned copy]
Eric Weisstein's World of Mathematics, Repunit.
Eric Weisstein's World of Mathematics, Weighing.


FORMULA

G.f.: x/((1x)*(13*x)).
a(n) = 4*a(n1)  3*a(n2), n > 1. a(0) = 0, a(1) = 1.
a(n) = 3*a(n1) + 1, a(0) = 0.
E.g.f.: (exp(3*x)  exp(x))/2.  Paul Barry, Apr 11 2003
a(n+1) = Sum_{k = 0..n} binomial(n+1, k+1)*2^k.  Paul Barry, Aug 20 2004
a(n) = Sum_{i = 0..n1} 3^i, for n > 0; a(0) = 0.
a(n) = StirlingS2(n+1, 3) + StirlingS2(n+1, 2).  Ross La Haye, Jan 10 2008
a(n) = 2*a(n1) + 3*a(n2) + 2, n > 1.  Gary Detlefs, Jun 21 2010
a(n) = 3*a(n1) + a(n2)  3*a(n3) = 5*a(n1)  7*a(n2) + 3*a(n3), a(0) = 0, a(1) = 1, a(2) = 4. Observation by G. Detlefs. See the W. Lang comment and link.  Wolfdieter Lang, Oct 18 2010
G.f.: Q(0)/2 where Q(k)= 1  1/(9^k  3*x*81^k/(3*x*9^k  1/(1  1/(3*9^k  27*x*81^k/(9*x*9^k  1/Q(k+1) ))))); (continued fraction ).  Sergei N. Gladkovskii, Apr 12 2013


EXAMPLE

There are 4 3block bicoverings of a 3set: {{1, 2, 3}, {1, 2}, {3}}, {{1, 2, 3}, {1, 3}, {2}}, {{1, 2, 3}, {1}, {2, 3}} and {{1, 2}, {1, 3}, {2, 3}}.
Ternary........Decimal
0.................0
1.................1
11................4
111..............13
There are altogether a(3) = 13 three letter words over {A,B,C} with say, A, appearing an odd number of times: AAA; ABC, ACB, ABB, ACC; BAC, CAB, BAB, CAC; BCA, CBA, BBA, CCA.  Wolfdieter Lang, Jul 16 2017


MAPLE



MATHEMATICA

LinearRecurrence[{4, 3}, {0, 1}, 30] (* Harvey P. Dale, Jul 13 2011 *)
CoefficientList[Series[x/(1  4x + 3x^2), {x, 0, 30}], x] (* Eric W. Weisstein, Sep 28 2017 *)
Table[FromDigits[PadRight[{}, n, 1], 3], {n, 0, 30}] (* Harvey P. Dale, Jun 01 2022 *)


PROG

(PARI) a(n)=(3^n1)/2
(Haskell)
a003462 = (`div` 2) . (subtract 1) . (3 ^)
(PARI) concat(0, Vec(x/((1x)*(13*x)) + O(x^30))) \\ Altug Alkan, Nov 01 2015
(GAP)


CROSSREFS

Cf. A064099 (minimal number of weightings to detect lighter or heavier coin among n coins).
Cf. A006516 (binomial transform, and special 4 letter words).


KEYWORD

nonn,easy,nice


AUTHOR



EXTENSIONS

Corrected my comment of Jan 10 2008.  Ross La Haye, Oct 29 2008
Removed comment that duplicated a formula.  Joerg Arndt, Mar 11 2010


STATUS

approved



