cp's OEIS Frontend

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

Also number of 3-block bicoverings of an n-set (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 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

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^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

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(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

It appears that this is A120444(3^n-1) = A004125(3^n) - A004125(3^n-1), 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, 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).

From Adi Dani, Jun 08 2011: (Start)

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

a(n) is the least number k such that A065363(k) = n. - Amiram Eldar, Sep 03 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

From Alexander Adamchuk, Jan 22 2007: (Start)

a(n) is divisible by (n-1).

Corresponding quotients are a(n)/(n-1) = {1,3,13,85,781,9331, ...} = A023037(n).

p divides a(p-1) for prime p.

p divides a((p-1)/2) for prime p = {3,11,17,19,41,43,59,67,73,83,89,97,...} = A033200 Primes congruent to {1, 3} mod 8; or, odd primes of form x^2+2*y^2.

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)/4) for prime p = {5,13,17,29,37,41,53,61,73,...} = A002144 Pythagorean primes: primes of form 4n+1.

p divides a((p-1)/5) for prime p = {31,191,251,271,601,641,761,1091,...}.

p divides a((p-1)/6) for prime p = {7,241,313,337,409,439,607,631,727,751,919,937,...}. (End)

For n > 1, a(n) is largest number that can be represented using n digits in the base-n number system. - Chinmaya Dash, Mar 31 2022

Primes p == 1 (mod 3) such that 2 is a cubic residue mod p.

Primes p == 1 (mod 6) such that 2 and -2 are both cubes (one implies the other) mod p. - Warren D. Smith

Subsequence of A040028, complement of A045309 relative to A040028. For p in this sequence, x^3 == 2 (mod p) has three solutions in integers from 0 to p-1, whose sum is p (A059899) or 2*p (A059914). The solutions are given in A060122, A060123 and A060124. - Klaus Brockhaus, Mar 02 2001

Primes p = 3m+1 such that 2^m == 1 (mod p). Subsequence of A016108 which also includes composites satisfying this congruence. - Alzhekeyev Ascar M, Feb 22 2012

a(n) is the smallest prime p == 1 (mod 3) such that each of the first n primes is a cubic residue mod p. - Robert Israel, Aug 02 2016

3 is a cube mod p for all primes in this list; this is a particular case of a result of Gauss. See Ireland and Rosen, Chapter 9, Exercise 23, p. 135. Some examples are given below.

Primes p such that 4*p - 243 is a square. Let p == 1 (mod 6) be a prime. There are integers c and d such that 4*p = c^2 + 27*d^2 (see, for example, Ireland and Rosen, Proposition 8.3.2). This sequence lists the primes with d = 3. Cf. A005471 (case d = 1) and A227622 (case d = 2).

Primes p of the form m^2 + 9*m + 81 are related to cyclic cubic fields in several ways:

(1) The cubic x^3 - p*x + 3*p, with discriminant ((2*m + 9)*p)^2, is irreducible over Q by Eisenstein's criteria. It follows that the Galois group of the polynomial over Q is the cyclic group C_3 (apply Conrad, Corollary 2.5).

Note that the roots of x^3 - p*x + 3*p are the differences n_0 - n_1, n_1 - n_2 and n_2 - n_0, where n_0, n_1 and n_2 are the three cubic Gaussian periods for the modulus p.

(2) The cubic x^3 - m*x^2 - 3*(m + 9)*x - 27 has discriminant (3*p)^2, a square. This is the polynomial g_3(a, 0, -3; X) in the notation of Hashimoto and Hoshi. The cubic is irreducible over Q by the Rational Root Theorem and hence the Galois group of the polynomial over Q is the cyclic group C_3.

(3) The cubic 3*x^3 + p*(x + 3)^2, with discriminant 81*p^2*(4*p - 243), a square, is irreducible over Q by Eisenstein's criteria. It follows that the Galois group of the polynomial over Q is the cyclic group C_3.