A066443 -id:A066443 - cp's OEIS frontend

A015518 a(n) = 2a(n-1) + 3a(n-2), with a(0)=0, a(1)=1.

0, 1, 2, 7, 20, 61, 182, 547, 1640, 4921, 14762, 44287, 132860, 398581, 1195742, 3587227, 10761680, 32285041, 96855122, 290565367, 871696100, 2615088301, 7845264902, 23535794707, 70607384120, 211822152361, 635466457082

Offset: 0

Olivier Gérard

Number of walks of length n between any two distinct vertices of the complete graph K_4. - Paul Barry and Emeric Deutsch, Apr 01 2004
For n >= 1, a(n) is the number of integers k, 1 <= k <= 3^(n-1), whose ternary representation ends in an even number of zeros (see A007417). - Philippe Deléham, Mar 31 2004
Form the digraph with matrix A=[0,1,1,1;1,0,1,1;1,1,0,1;1,0,1,1]. A015518(n) corresponds to the (1,3) term of A^n. - Paul Barry, Oct 02 2004
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is 2. - Cino Hilliard, Sep 25 2005
(A046717(n))^2 + (2*a(n))^2 = A046717(2n). E.g., A046717(3) = 13, 2*a(3) = 14, A046717(6) = 365. 13^2 + 14^2 = 365. - Gary W. Adamson, Jun 17 2006
For n >= 2, number of ordered partitions of n-1 into parts of sizes 1 and 2 where there are two types of 1 (singletons) and three types of 2 (twins). For example, the number of possible configurations of families of n-1 male (M) and female (F) offspring considering only single births and twins, where the birth order of M/F/pair-of-twins is considered and there are three types of twins; namely, both F, both M, or one F and one M - where birth order within a pair of twins itself is disregarded. In particular, for a(3)=7, two children could be either: (1) F, then M; (2) M, then F; (3) F,F; (4) M,M; (5) F,F twins; (6) M,M twins; or (7) M,F twins (emphasizing that birth order is irrelevant here when both/all children are the same gender and when two children are within the same pair of twins). - Rick L. Shepherd, Sep 18 2004
a(n) is prime for n = {2, 3, 5, 7, 13, 23, 43, 281, 359, ...}, where only a(2) = 2 corresponds to a prime of the form (3^k - 1)/4. All prime terms, except a(2) = 2, are the primes of the form (3^k + 1)/4. Numbers k such that (3^k + 1)/4 is prime are listed in A007658. Note that all prime terms have prime indices. Prime terms are listed in A111010. - Alexander Adamchuk, Nov 19 2006
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=charpoly(A,1). - Milan Janjic, Jan 26 2010
Select an odd size subset S from {1,2,...,n}, then select an even size subset from S. - Geoffrey Critzer, Mar 02 2010
a(n) is the number of ternary sequences of length n where the numbers of (0's, 1's) are (even, odd) respectively, and, by symmetry, the number of such sequences where those numbers are (odd, even) respectively. A122983 covers (even, even), and A081251 covers (odd, odd). - Toby Gottfried, Apr 18 2010
An elephant sequence, see A175654. For the corner squares just one A[5] vector, with decimal value 341, leads to this sequence (without the leading 0). For the central square this vector leads to the companion sequence A046717 (without the first leading 1). - Johannes W. Meijer, Aug 15 2010
Let R be the commutative algebra resulting from adjoining the elements of the Klein four-group to the integers (equivalently, K = Z[x,y,z]/{x*y - z, y*z - x, x*z - y, x^2 - 1, y^2 - 1, z^2 - 1}). Then a(n) is equal to the coefficients of x, y, and z in the expansion of (x + y + z)^n. - Joseph E. Cooper III (easonrevant(AT)gmail.com), Nov 06 2010
Pisano period lengths:  1, 2, 2, 4, 4, 2, 6, 8, 2, 4, 10, 4, 6, 6, 4, 16, 16, 2, 18, 4, ... - R. J. Mathar, Aug 10 2012
The ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 09 2014
This is a divisibility sequence, also the values of Chebyshev polynomials, and also the number of ways of packing a 2 X n-1 rectangle with dominoes and unit squares. - R. K. Guy, Dec 16 2016
For n>0, gcd(a(n),a(n+1))=1. - Kengbo Lu, Jul 02 2020

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
Jean-Paul Allouche, Jeffrey Shallit, Zhixiong Wen, Wen Wu, and Jiemeng Zhang, Sum-free sets generated by the period-k-folding sequences and some Sturmian sequences, arXiv:1911.01687 [math.CO], 2019.
Katerina Böhmová, Cristina Dalfó, and Clemens Huemer, On cyclic Kautz digraphs, Research Report, 2015.
Garry Bowlin and Matthew G. Brin, Coloring Planar Graphs via Colored Paths in the Associahedra, arXiv preprint arXiv:1301.3984 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 12 2013
AJ Bu and Doron Zeilberger, Using Symbolic Computation to Explore Generalized Dyck Paths and Their Areas, arXiv:2305.09030 [math.CO], 2023.
Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
Sergio Falcón, Binomial Transform of the Generalized k-Fibonacci Numbers, Communications in Mathematics and Applications (2019) Vol. 10, No. 3, 643-651.
Dale Gerdemann, Fractal generated from (2,3) recursion, YouTube Video, Dec 05 2014.
Toufik Mansour and Mark Shattuck, Pattern avoidance in flattened derangements, Disc. Math. Lett. (2025) Vol. 15, 67-74. See p. 74.
R. J. Mathar, Counting Walks on Finite Graphs, Nov 2020, Section 2.
Felix Pozon Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014.
Index entries for linear recurrences with constant coefficients, signature (2,3).
Index entries for sequences related to Chebyshev polynomials.

a(n) = A080926(n-1) + 1 = (1/3)*A054878(n+1) = (1/3)*abs(A084567(n+1)).
First differences of A033113 and A039300.
Partial sums of A046717.
The following sequences (and others) belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A063727, A083098, A083099, A083100, A084057.
Cf. A046717.
Cf. A007658, A111010.
Cf. A001045, A078008, A097073, A115341. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008
Cf. A007482, A059260.

[Round(3^n/4): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011

Table[(3^n-(-1)^n)/4,{n,0,30}] (* Alexander Adamchuk, Nov 19 2006 *)

a(n):= round(3^n/4)$ /* Dimitri Papadopoulos, Nov 28 2023 */

PARI
```
a(n)=round(3^n/4)
```

for n in range(0, 20): print(int((3**n-(-1)**n)/4), end=', ') # Stefano Spezia, Nov 30 2018

Sage
```
[round(3^n/4) for n in range(0,27)]
```

G.f.: x/((1+x)*(1-3*x)).
a(n) = (3^n - (-1)^n)/4 = floor(3^n/4 + 1/2).
a(n) = 3^(n-1) - a(n-1). - Emeric Deutsch, Apr 01 2004
E.g.f.: (exp(3*x) - exp(-x))/4. Second inverse binomial transform of (5^n-1)/4, A003463. Inverse binomial transform for powers of 4, A000302 (when preceded by 0). - Paul Barry, Mar 28 2003
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k+1)*2^(2k). - Paul Barry, May 14 2003
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*4^(k-1). - Paul Barry, Apr 02 2003
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2*k)*3^k. - Paul Barry, Jul 13 2004
a(n) = U(n-1, i/sqrt(3))(-i*sqrt(3))^(n-1), i^2=-1. - Paul Barry, Nov 17 2003
G.f.: x*(1+x)^2/(1 - 6*x^2 - 8*x^3 - 3*x^4) = x(1+x)^2/characteristic polynomial(x^4*adj(K_4)(1/x)). - Paul Barry, Feb 03 2004
a(n) = sum_{k=0..3^(n-1)} A014578(k) = -(-1)^n*A014983(n) = A051068(3^(n-1)), for n > 0. - Philippe Deléham, Mar 31 2004
E.g.f.: exp(x)*sinh(2*x)/2. - Paul Barry, Oct 02 2004
a(2*n+1) = A054880(n) + 1. - M. F. Hasler, Mar 20 2008
2*a(n) + (-1)^n = A046717(n). - M. F. Hasler, Mar 20 2008
a(n) = ((1+sqrt(4))^n - (1-sqrt(4))^n)/4. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008
a(n) = abs(A014983(n)). - Zerinvary Lajos, May 28 2009
a(n) = round(3^n/4). - Mircea Merca, Dec 28 2010
a(n) = Sum_{k=1,3,5,...} binomial(n,k)*2^(k-1). - Geoffrey Critzer, Mar 02 2010
From Sergei N. Gladkovskii, Jul 19 2012: (Start)
G.f.: G(0)/4 where G(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/G(k+1)))))); (continued fraction).
E.g.f.: G(0)/4 where G(k)= 1 - 1/(9^k - 3*x*81^k/(3*x*9^k - (2*k+1)/(1 + 1/(3*9^k - 27*x*81^k/(9*x*9^k + (2*k+2)/G(k+1)))))); (continued fraction).  (End)
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(4*k-1)/(x*(4*k+3) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n+1) = Sum_{k = 0..n} A238801(n,k)*2^k. - Philippe Deléham, Mar 07 2014
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-4)^k = (-1)^(n-1)*Sum_{k=0..n-1} (-3)^k. Equals (-1)^(n-1)*Phi(n,-3), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014
a(n) = 2*A006342(n-1) - n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Nov 30 2018
a(n) = 2*A033113(n-2) + n mod 2 if n > 0, a(0)=0. - Yuchun Ji, Aug 16 2019
a(2*k) = 2*A002452(k), a(2*k+1) = A066443(k). - Yuchun Ji, Aug 14 2019
a(n+1) = 2*Sum_{k=0..n} a(k) if n odd, and 1 + 2*Sum_{k=0..n} a(k) if n even. - Kengbo Lu, May 30 2020
a(n) = F(n) + Sum_{k=1..(n-1)} a(k)*L(n-k), for F(n) and L(n) the Fibonacci and Lucas numbers. - Kengbo Lu and Greg Dresden, Jun 05 2020
From Kengbo Lu, Jun 11 2020: (Start)
a(n) = A002605(n) + Sum_{k = 1..n-2} a(k)*A002605(n-k-1).
a(n) = A006130(n-1) + Sum_{k = 1..n-1} a(k)*A006130(n-k-1). (End)
a(2n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)* 2^(2n-2i-2j-1)* 3^(i+j). - Kengbo Lu, Jul 02 2020
a(n) = 3*a(n-1) - (-1)^n. - Dimitri Papadopoulos, Nov 28 2023
G.f.: x/((1 + x)*(1 - 3*x)) = Sum_{n >= 0} x^(n+1) * Product_{k = 1..n} (k + 3*x + 1)(1 + k*x) (a telescoping series). Cf. A007482. - Peter Bala, May 08 2024
From Peter Bala, Jun 29 2025: (Start)
For n >= 1, a(n+1) = 2^n * hypergeom([1/2 - (1/2)*n, -(1/2)*n], [-n], -3).
G.f. A(x) = x*exp(Sum_{n >= 1} a(2*n)/a(n)*x^n/n) =  x + 2*x^2 + 7*x^3 + 20*x^4 + ....
sqrt(A(x)/x) is the g.f. of A002426.
The following series telescope:
Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)) = -1; Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*a(n+2)*a(n+3)) = -1/98.
In general, for k >= 0, Sum_{n >= 1} (-3)^n/(a(n)*a(n+1)*...*a(n+2*k+1)) = -1/((a(1)*a(2)*...*a(2*k+1))*a(2*k+1)).
Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)) = 1/4; Sum_{n >= 1} 3^n/(a(n)*a(n+1)*a(n+2)* a(n+3)*a(n+4)) = 1/5600.
In general, for k >= 1, Sum_{n >= 1} 3^n/(a(n)*a(n+1)*...*a(n+2*k)) = 1/((a(1)*a(2)*...*a(2*k))*a(2*k)). (End)

More terms from Emeric Deutsch, Apr 01 2004

Edited by Ralf Stephan, Aug 30 2004

A083233 a(n) = (3*8^n + 0^n)/4.

Original entry on oeis.org

1, 6, 48, 384, 3072, 24576, 196608, 1572864, 12582912, 100663296, 805306368, 6442450944, 51539607552, 412316860416, 3298534883328, 26388279066624, 211106232532992, 1688849860263936, 13510798882111488, 108086391056891904, 864691128455135232

Offset: 0

Paul Barry, Apr 23 2003

Binomial transform of A083232. Inverse binomial transform of A066443.
Numbers k such that, except for some first term, k^2 = [A000302]^3 + [A004171]^3 + [A002001]^3; e.g., 3072^2 = 64^3 + 128^3 + 192^3; 51539607552^2 = 4194304^3 + 8388608^3 + 12582912^3. - Vincenzo Librandi, Aug 08 2010
With the exception of the first term, these numbers cannot be written as the sum of two integer cubes but can be written as the sum of two positive rational cubes (i.e., 6*8^n = (17*2^n/21)^3 + (37*2^n/21)^3). - Arkadiusz Wesolowski, Aug 15 2013
a(n+1) is the number of unit square faces on the convex hull of a level n Menger sponge. This follows since it has six exterior faces, each of which is a Sierpinski carpet with 8^n squares. - Allan Bickle, Nov 28 2022

			a(0) = (3*8^0 + 0^0)/4 = 4/4 = 1 (using 0^0 = 1).

Allan Bickle, Degrees of Menger and Sierpinski Graphs, Congr. Num. 227 (2016) 197-208.
Allan Bickle, MegaMenger Graphs, The College Mathematics Journal, 49 1 (2018) 20-26.
Eric Weisstein's World of Mathematics, Menger Sponge
Wikipedia, Menger sponge
Index entries for linear recurrences with constant coefficients, signature (8).

Cf. A083234. Subsequence of A159843.
Cf. A000302, A004171, A002001.
Cf. A291066, A083233, and A332705 on the surface area of the n-Menger sponge graph.

Join[{1},NestList[8#&,6,20]] (* Harvey P. Dale, Sep 25 2020 *)

a(n)=(3*8^n+0^n)/4 \\ Charles R Greathouse IV, Oct 07 2015

def A083233(n): return 3<<3*n-2 if n else 1 # Chai Wah Wu, Nov 27 2023

a(n) = (3*8^n + 0^n)/4.
G.f.: (1-2x)/(1-8x).
E.g.f.: (3*exp(8x) + exp(0))/4.
a(0) = 1, a(n+1) = 6*8^n. - Arkadiusz Wesolowski, Aug 15 2013

A122983 a(n) = (2 + (-1)^n + 3^n)/4.

Original entry on oeis.org

1, 1, 3, 7, 21, 61, 183, 547, 1641, 4921, 14763, 44287, 132861, 398581, 1195743, 3587227, 10761681, 32285041, 96855123, 290565367, 871696101, 2615088301, 7845264903, 23535794707, 70607384121, 211822152361, 635466457083

Offset: 0

Paul Barry, Sep 22 2006

Old definition was: "Binomial transform of aeration of A081294".
Binomial transform is A063376.
A122983 = (1,1,3,7,1,1,3,7,...) mod 10. - M. F. Hasler, Feb 25 2008
Equals row sums of triangle A158301. - Gary W. Adamson, Mar 15 2009
a(n) = the number of ternary sequences of length n where the numbers of (0's, 1's) are both even. A015518 covers the (odd, even) and (even, odd) cases, and A081251 covers (odd, odd). - Toby Gottfried, Apr 18 2010
This sequence also describes the number of moves of the k-th disk solving (non-optimally) the [RED ; NEUTRAL ; BLUE] pre-colored Magnetic Tower of Hanoi (MToH) puzzle. The sequence A183119 is the partial sums of the sequence in question (obviously describing the total number of moves associated with the specific solution algorithm). For other MToH-related sequences, Cf. A183111 - A183125.
Let B=[1,sqrt(2),0; sqrt(2),1,sqrt(2); 0,sqrt(2),1] be a 3 X 3 matrix. Then a(n)=[B^n](1,1), n=0,1,2,.... - _L. Edson Jeffery, Dec 21 2011
Also the domination number of the n-Hanoi graph. - Eric W. Weisstein, Jun 16 2017
Also the matching number of the n-Sierpinski gasket graph. - Eric W. Weisstein, Jun 17 2017
Let M = [1,1,1,0; 1,1,0,1; 1,0,1,1; 0,1,1,1], a 4 X 4 matrix. Then a(n) is the upper left entry in M^n. - Philippe Deléham, Aug 23 2020
Also the lower matching number (=independent domination number) of the n-Hanoi graph. - Eric W. Weisstein, Aug 01 2023

M. F. Hasler, Table of n, a(n) for n = 0..199.
Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
Alexander Diaz-Lopez, Pamela E. Harris, Erik Insko, and Darleen Perez-Lavin, Peaks Sets of Classical Coxeter Groups, arXiv preprint, arXiv:1505.04479 [math.GR], 2015.
A. M. Hinz, S. Klavžar, U. Milutinović, and C. Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 99. Book's website
Uri Levy, The Magnetic Tower of Hanoi, arXiv:1003.0225 [math.CO], 2010.
Eric Weisstein's World of Mathematics, Domination Number.
Eric Weisstein's World of Mathematics, Hanoi Graph.
Eric Weisstein's World of Mathematics, Lower Independence Number.
Eric Weisstein's World of Mathematics, Matching Number.
Eric Weisstein's World of Mathematics, Sierpiński Gasket Graph.
Index entries for linear recurrences with constant coefficients, signature (3,1,-3).

Cf. a(j+1) = A137822(2^j) and these are the record values of A137822.

Cf. A054879 (bisection), A066443 (bisection). Row sums of A158303.

A122983 := n -> ceil(3^n/4); 'A122983(n)' $ n=0..22; # M. F. Hasler, Feb 25 2008
a[ -1]:=1:a[0]:=1:a[1]:=3:for n from 2 to 50 do a[n]:=2*a[n-1]+3*a[n-2]-2 od: seq(a[n], n=-1..25); # Zerinvary Lajos, Apr 28 2008

CoefficientList[Series[(1 - 2 x - x^2)/((1 - x) (1 + x) (1 - 3 x)), {x, 0, 40}], x] (* Harvey P. Dale, Sep 03 2013 *)
LinearRecurrence[{3, 1, -3}, {1, 1, 3}, 40] (* Harvey P. Dale, Sep 03 2013 *)
Table[(2 + (-1)^n + 3^n)/4, {n, 0, 20}] (* Eric W. Weisstein, Jun 16 2017 *)
Table[Floor[3^n/4] + 1, {n, 0, 20}] (* Eric W. Weisstein, Jan 17 2018 *)
Floor[3^Range[0, 20]/4] + 1 (* Eric W. Weisstein, Jan 17 2018 *)

A122983(n)=3^n\4+1 \\ M. F. Hasler, Feb 25 2008

def A122983(n): return (1 if n&1 else 3)+3**n>>2 # Chai Wah Wu, Apr 12 2023

From Paul Barry, Jun 14 2007: (Start)
G.f.: (1-2*x-x^2)/((1-x)*(1+x)*(1-3*x));
a(n) = 3^n/4+(-1)^n/4+1/2;
E.g.f.: cosh(x)^2*exp(x). (End)
a(n) = 3*a(n-1) + a(n-2) - 3*a(n-3); a(0)=1, a(1)=1, a(2)=3. - Harvey P. Dale, Sep 03 2013
E.g.f.: Q(0)/2, where Q(k) = 1 + 3^k/( 2 - 2*(-1)^k/( 3^k + (-1)^k - 2*x*3^k/( 2*x + (k+1)*(-1)^k/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Dec 22 2013
a(2*n) = 3*a(2*n-1); a(2*n+1) = 3*a(2*n) - 2. - Philippe Deléham, Aug 23 2020

Extended and corrected (existing Maple code) by M. F. Hasler, Feb 25 2008

Description changed to formula by Eric W. Weisstein, Jun 16 2017

A083075 Square array read by antidiagonals: T(n,k) = (k(2k+3)^n + 1)/(k+1).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 13, 1, 1, 7, 33, 63, 1, 1, 9, 61, 229, 313, 1, 1, 11, 97, 547, 1601, 1563, 1, 1, 13, 141, 1065, 4921, 11205, 7813, 1, 1, 15, 193, 1831, 11713, 44287, 78433, 39063, 1, 1, 17, 253, 2893, 23801, 128841, 398581, 549029, 195313, 1, 1, 19, 321

Offset: 0

Paul Barry, Apr 23 2003

			Array begins:
  1     1     1     1     1 ...
  1     3    13    63   313 ...
  1     5    33   229  1601 ...
  1     7    61   547  4921 ...
  1     9    97  1065 11713 ...
  ...

Nathaniel Johnston, Antidiagonals 0..100, flattened

Rows include A034478, A083076, A066443, A083077, A083078.
Columns include odds, A082109, A083079.
Diagonals include A083079, A083080, A083081, A083082.

T := proc(n,k) return (k*(2*k+3)^n+1)/(k+1): end: seq(seq(T(k,n-k),k=0..n),n=0..10); # Nathaniel Johnston, Jun 26 2011

A102591 a(n) = Sum_{k=0..n} binomial(2n+1, 2k)*3^(n-k).

Original entry on oeis.org

1, 6, 44, 328, 2448, 18272, 136384, 1017984, 7598336, 56714752, 423324672, 3159738368, 23584608256, 176037912576, 1313964867584, 9807567290368, 73204678852608, 546407161659392, 4078438577864704, 30441879976280064

Offset: 0

Paul Barry, Jan 22 2005

In general, Sum_{k=0..n} binomial(2n+1,2k)*r^(n-k) has g.f. (1-(r-1)x)/(1-2(r+1)+(r-1)^2x^2) and a(n) = ((sqrt(r)-1)^(2n+1) + (sqrt(r)+1)^(2n+1))/(2*sqrt(r)).

Index entries for linear recurrences with constant coefficients, signature (8,-4).

Bisection of A080879 or A002605.

Cf. A066443, A079935, A099156, A102592, A107903.

LinearRecurrence[{8,-4},{1,6},20] (* Harvey P. Dale, Sep 28 2021 *)

G.f.: (1-2x)/(1-8x+4x^2);
a(n) = 8*a(n-1) - 4*a(n-2);
a(n) = sqrt(3)*(sqrt(3)-1)^(2n+1)/6 + sqrt(3)*(sqrt(3)+1)^(2n+1)/6.
a(n) = 2^n*A079935(n). - R. J. Mathar, Sep 20 2012
a(n) = 2^(2*n+1)*Sum_{k >= n} binomial(2*k,2*n)*(1/3)^(k+1). Cf. A099156. - Peter Bala, Nov 29 2021
3*a(n)^2 = A107903(n)^2 + 2^(2*n+1). - Philippe Deléham, Mar 21 2023

A083076 Third row of number array A083075.

Original entry on oeis.org

1, 5, 33, 229, 1601, 11205, 78433, 549029, 3843201, 26902405, 188316833, 1318217829, 9227524801, 64592673605, 452148715233, 3165041006629, 22155287046401, 155087009324805, 1085609065273633, 7599263456915429, 53194844198408001

Offset: 0

Paul Barry, Apr 23 2003

Binomial transform of A067411. Inverse binomial transform of A082412.

Trinomial transform of Jacobsthal numbers A001045. - Paul Barry, Sep 10 2007

Nathaniel Johnston, Table of n, a(n) for n = 0..250
Index entries for linear recurrences with constant coefficients, signature (8,-7).

Cf. A034478, A066443, A082761, A083077.

[(2*7^n+1)/3: n in [0..30]]; // Vincenzo Librandi, Nov 06 2011

seq((2*7^n+1)/3,n=0..20); # Nathaniel Johnston, Jun 26 2011

a(n) = (2*7^n + 1)/3.
G.f.: (1-3*x)/((1-x)*(1-7*x)).
E.g.f.: (2*exp(7*x) + exp(x))/3.
a(n) = Sum_{k=0..2*n} trinomial(n,k)*Fibonacci(k+1), where trinomial(n,k) are the trinomial coefficients (A027907). - Paul Barry, Sep 10 2007
a(n) = 7*a(n-1) - 2, a(n) = 8*a(n-1) - 7*a(n-2). - Vincenzo Librandi, Nov 06 2011

A322469 Irregular table: row i = 1, 2, 3, ... starts with 4*i - 1; then, as long as the number is divisible by 3, the next two terms are the result of dividing it by 3, then multiplying it by 2.

Original entry on oeis.org

3, 1, 2, 7, 11, 15, 5, 10, 19, 23, 27, 9, 18, 6, 12, 4, 8, 31, 35, 39, 13, 26, 43, 47, 51, 17, 34, 55, 59, 63, 21, 42, 14, 28, 67, 71, 75, 25, 50, 79, 83, 87, 29, 58, 91, 95, 99, 33, 66, 22, 44, 103, 107, 111, 37, 74

Offset: 1

Georg Fischer, Dec 09 2018

The sequence is the flattened form of an irregular table T(i, j) (see the example below) which has rows i >= 1 consisting of subsequences of varying length as defined by the following algorithm:
    j := 1; T(i, j) := 4 * i - 1;
    while T(i, j) is divisible by 3 do
        T(i, j + 1) := T(i, j) / 3;
        T(i, j + 2) := T(i, j + 1) * 2;
        j := j + 2;
    end while
The algorithm always stops.
The first rows which are longer than any previous row are 1, 7, 61, 547, 4921 ... (A066443).
Property: The sequence is a permutation of the natural numbers > 0.
Proof: (Start)
The values in the columns j of T for row indexes i of the form i = e * k + f,
k >= 0, if such columns are present, have the following residues modulo some power of 2:
j | Op.  | Form of i    |  T(i, j)     |  Residues  | Residues not yet covered
--+------+ -------------+--------------+------------+-------------------------
1 |      |  1 * k +  1  |   4 * k +  3 |   3 mod  4 |   0,  1,  2     mod  4
2 | / 3  |  3 * k +  1  |   4 * k +  1 |   1 mod  4 |   0,  2,  4,  6 mod  8
3 | * 2  |  3 * k +  1  |   8 * k +  2 |   2 mod  8 |   0,  4,  6     mod  8
4 | / 3  |  9 * k +  7  |   8 * k +  6 |   6 mod  8 |   0,  4,  8, 12 mod 16
5 | * 2  |  9 * k +  7  |  16 * k + 12 |  12 mod 16 |   0,  4,  8     mod 16
6 | / 3  | 27 * k +  7  |  16 * k +  4 |   4 mod 16 |   0,  8, 16, 24 mod 32
7 | * 2  | 27 * k +  7  |  32 * k +  8 |   8 mod 32 |   0, 16, 24     mod 32
8 | / 3  | 81 * k + 61  |  32 * k + 24 |  24 mod 32 |   0, 16, 32, 48 mod 64
9 | * 2  | 81 * k + 61  |  64 * k + 48 |  48 mod 64 |   0, 16, 32     mod 64
..| ...  |  e * k +  f  |   g * k +  m |   m mod  g |   0, ...
The variables in the last, general line can be computed from the operations in the algorithm. They are the following:
  e = 3^floor(j / 2)
  f = A066443(floor(j / 4)) with A066443(n) = (3^(2*n+1)+1)/4
  g = 2^floor((j + 3) / 2)
  m = 2^floor((j - 1) / 4) * A084101(j + 1 mod 4) with A084101(0..3) = (1, 3, 3, 1)
The residues m in each column and therefore the T(i, j) are all disjoint. For numbers which contain a sufficiently high power of 3, the length of the rows in T grows beyond any limit, and the numbers containing any power of 2 will finally be covered.
(End)
All numbers > 0 up to and including 2^(2*j + 1) appear in the rows in T up to and including A066443(j). For example, 4096 and 8192 are the trailing elements in row 398581 = A066443(6).
Length of row n = 1, 2, ... is 1+2*A007949(A004767(n-1)). - M. F. Hasler, Dec 10 2018
From Georg Fischer, Oct 16 2020: (Start)
Whenever a row of T is longer than any previous rows, it defines the start values of the arithmetic progressions in the additional columns. These start values form the sequence A308709.
There is a hierarchy of such permutations of the positive integers derived by selecting and mapping the terms of the form 6*k - 2 to k:
  Level 0: A307407, nodes in the graph of the "3x+1" or Collatz problem
  Level 1: A322469 (this sequence), inverse is A338208
  Level 2: A307048, inverse is A338207
  Level 3: A160016, inverse is A338206
  Level >= 4: A000027, the positive integers
Conjectures (verified for k = 0..11):
a(A338186(k)) = 4^k.
If A338186(k) <= j < A338186(k+1) then a(A338186(k)) <= a(j).
(End)

			Table T(i, j) begins:
  i\j   1  2  3  4  5  6  7
  -------------------------
  1:    3  1  2
  2:    7
  3:   11
  4:   15  5 10
  5:   19
  6:   23
  7:   27  9 18  6 12  4  8

Alois P. Heinz, Rows n = 1..10000, flattened
Index entries for sequences that are permutations of the positive integers

Cf. A066443, A084101, A160016 (level 3), A307048 (level 2), A307407 (level 0), A308709, A338186, A338206, A338207, A338208.

T:= proc(n) local m, l; m:= 4*n-1; l:= m;
      while irem(m, 3, 'm')=0 do
         l:= l, m; m:= m*2; l:=l, m;
      od; l
    end:
seq(T(n), n=1..40);  # Alois P. Heinz, Dec 10 2018

s={}; Do[a=4n-1; AppendTo[s,a]; While[Divisible[a, 3], a/=3; AppendTo[s, a]; a*=2; AppendTo[s, a]], {n, 1, 30}]; s (* Amiram Eldar, Dec 10 2018 *)

apply( A322469_row(n,L=[n=4*n+3])={while(n%3==0,L=concat(L,[n\=3, n*=2]));L}, [0..99]) \\ Use concat(%) to flatten the table if desired. - M. F. Hasler, Dec 10 2018

use integer; my $n = 1; my $i = 1;
  while ($i <= 1000) { # next row
    my $an = 4 * $i - 1; print "$n $an\n"; $n ++;
    while ($an % 3 == 0) {
      $an /= 3; print "$n $an\n"; $n ++;
      $an *= 2; print "$n $an\n"; $n ++;
    } # while divisible by 3
    $i ++;
} # while next row - Georg Fischer, Dec 12 2018

def A322469_list(len):
    L = []
    for n in (1..len):
        a = 4*n - 1
        L.append(a)
        while 3.divides(a):
            a //= 3
            L.append(a)
            a <<= 1
            L.append(a)
    return L
A322469_list(28) # Peter Luschny, Dec 10 2018

A054880 a(n) = 3*(9^n - 1)/4.

Original entry on oeis.org

0, 6, 60, 546, 4920, 44286, 398580, 3587226, 32285040, 290565366, 2615088300, 23535794706, 211822152360, 1906399371246, 17157594341220, 154418349070986, 1389765141638880, 12507886274749926, 112570976472749340, 1013138788254744066, 9118249094292696600, 82064241848634269406, 738578176637708424660

Offset: 0

Paolo Dominici (pl.dm(AT)libero.it), May 23 2000

Number of walks of length 2n+1 along the edges of a (3 dimensional) cube between two opposite vertices.

Urn A initially contains 3 labeled balls while urn B is empty. A ball is randomly selected and switched from one urn to the other. a(n)/3^(2n+1) is the probability that urn A is empty after 2n+1 switches. - Geoffrey Critzer, May 23 2013

G. C. Greubel, Table of n, a(n) for n = 0..1000
G. Benkart and D. Moon, A Schur-Weyl Duality Approach to Walking on Cubes, arXiv preprint arXiv:1409.8154 [math.RT], 2014 and Ann. Combin. 20 (3) (2016) 397-417
R. J. Mathar, Counting Walks on Finite Graphs, Nov 2020, Section 5.
Index entries for linear recurrences with constant coefficients, signature (10,-9).

Cf. A002452, A015518, A046717, A066443.

List([0..30], n-> 3*(9^n -1)/4); # G. C. Greubel, Jul 14 2019

[3*(9^n -1)/4: n in [0..30]]; // G. C. Greubel, Jul 14 2019

Table[(2 n + 1)! Coefficient[Series[Sinh[x]^3, {x, 0, 2 n + 1}],
x^(2 n + 1)], {n, 0, 30}]  (* Geoffrey Critzer, May 23 2013 *)
LinearRecurrence[{10,-9},{0,6},30] (* Harvey P. Dale, Sep 17 2024 *)

vector(30, n, n--; 3*(9^n -1)/4) \\ G. C. Greubel, Jul 14 2019

[3*(9^n -1)/4 for n in (0..30)] # G. C. Greubel, Jul 14 2019

G.f.: (3/4)/(1 - 9*x) - (3/4)/(1 - x).
a(n) = 6*A002452(n).
sin(x)^3 = Sum_{k>=0} (-1)^(k+1)*a(k)*x^(2k+1)/(2k+1)!. - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 08 2001
a(n) = A015518(2n+1) - 1 = (A046717(2n+1) - 1)/2. - M. F. Hasler, Mar 20 2008
a(n) = 9*a(n-1) + 6 with n > 0, a(0) = 0. - Vincenzo Librandi, Aug 07 2010
a(n) = A066443(n) - 1. - Georg Fischer, Nov 25 2018
E.g.f.: 3*(exp(9*x) - exp(x))/4. - G. C. Greubel, Jul 14 2019
a(n) = 10*a(n-1) - 9*a(n-2) with a(0) = 0 and a(1) = 6. - Miquel A. Fiol, Mar 09 2024

cp's OEIS Frontend

A015518 a(n) = 2*a(n-1) + 3*a(n-2), with a(0)=0, a(1)=1.

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A083233 a(n) = (3*8^n + 0^n)/4.

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A122983 a(n) = (2 + (-1)^n + 3^n)/4.

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A083075 Square array read by antidiagonals: T(n,k) = (k*(2*k+3)^n + 1)/(k+1).

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A102591 a(n) = Sum_{k=0..n} binomial(2n+1, 2k)*3^(n-k).

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A083076 Third row of number array A083075.

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A322469 Irregular table: row i = 1, 2, 3, ... starts with 4*i - 1; then, as long as the number is divisible by 3, the next two terms are the result of dividing it by 3, then multiplying it by 2.

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A015518 a(n) = 2a(n-1) + 3a(n-2), with a(0)=0, a(1)=1.

A083075 Square array read by antidiagonals: T(n,k) = (k(2k+3)^n + 1)/(k+1).