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A003462
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a(n) = (3^n - 1)/2.
(Formerly M3463)
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289
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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
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0,3
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COMMENTS
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Partial sums of A000244. Values of base 3 strings of 1's.
a(n) = (3^n-1)/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 = 1-x 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^n-1)/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(i-1)| = 1 for i = 1, 2, ..., 2*n + 1, s(0) = 1, s(2n+1) = 4. - Herbert Kociemba, Jun 10 2004
Number of non-degenerate right-angled incongruent integer-edged 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((p-1)/5) for prime p = {41, 431, 491, 661, 761, 1021, 1051, 1091, 1171, ...}. p divides a((p-1)/4) for prime p = {13, 109, 181, 193, 229, 277, 313, 421, 433, 541, ...}. p divides a((p-1)/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((p-1)/2) for prime p = {11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, ...} = A097933(n). p divides a(p-1) for prime p > 7. p^2 divides a(p*(p-1)k) for all prime p except p = 3. p^3 divides a(p*(p-1)*(p-2)k) for prime p = 11. - Alexander Adamchuk, Jan 22 2007
Let P(A) be the power set of an n-element 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 2-combinations". - 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^n-1)/2 = a(n). - Chayim Lowen, Mar 03 2015]
Also, still with P(A) being the power set of a n-element set A, a(n) is the number of 2-element 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, i-1] = -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(n-1)+1)/(s(n-1)+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] pre-colored 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 n-th 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 n-Apollonian network. - Andrew Howroyd, Sep 02 2017
For n > 1, the number of triangles (cliques of size 3) in the (n-1)-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 nesting-stars, 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(n-1) + 1) create Sierpinski-anti-triangles, 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
a(n) is also the number of 3-cycles in the n-Dorogovtsev-Goltsev-Mendes graph (using the convention the 0-Dorogovtsev-Goltsev-Mendes graph is P_2). - Eric W. Weisstein, Dec 06 2023
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REFERENCES
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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, Springer-Verlag, NY, 2nd ed., 1989, p. 60.
Paulo Ribenboim, The Little Book of Big Primes, Springer-Verlag, NY, 1991, p. 53.
Amir Sapir, The Tower of Hanoi with Forbidden Moves, The Computer J. 47 (1) (2004) 20, case three-in-a 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).
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LINKS
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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. 65-79. ISBN 978-0-691-16403-8
G. Kreweras, Sur les éventails de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #15 (1970), 3-41. [Annotated scanned copy]
Eric Weisstein's World of Mathematics, Repunit.
Eric Weisstein's World of Mathematics, Weighing.
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FORMULA
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G.f.: x/((1-x)*(1-3*x)).
a(n) = 4*a(n-1) - 3*a(n-2), n > 1. a(0) = 0, a(1) = 1.
a(n) = 3*a(n-1) + 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..n-1} 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(n-1) + 3*a(n-2) + 2, n > 1. - Gary Detlefs, Jun 21 2010
a(n) = 3*a(n-1) + a(n-2) - 3*a(n-3) = 5*a(n-1) - 7*a(n-2) + 3*a(n-3), 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
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EXAMPLE
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There are 4 3-block bicoverings of a 3-set: {{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
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MAPLE
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MATHEMATICA
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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 *)
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PROG
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(PARI) a(n)=(3^n-1)/2
(Haskell)
a003462 = (`div` 2) . (subtract 1) . (3 ^)
(PARI) concat(0, Vec(x/((1-x)*(1-3*x)) + O(x^30))) \\ Altug Alkan, Nov 01 2015
(GAP)
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CROSSREFS
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Cf. A064099 (minimal number of weightings to detect lighter or heavier coin among n coins).
Cf. A006516 (binomial transform, and special 4 letter words).
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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Corrected my comment of Jan 10 2008. - Ross La Haye, Oct 29 2008
Removed comment that duplicated a formula. - Joerg Arndt, Mar 11 2010
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STATUS
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approved
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