A017617 -id:A017617 - cp's OEIS frontend

A016789 a(n) = 3*n + 2.

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179

Offset: 0

N. J. A. Sloane

Except for 1, n such that Sum_{k=1..n} (k mod 3)*binomial(n,k) is a power of 2. - Benoit Cloitre, Oct 17 2002
The sequence 0,0,2,0,0,5,0,0,8,... has a(n) = n*(1 + cos(2*Pi*n/3 + Pi/3) - sqrt(3)*sin(2*Pi*n + Pi/3))/3 and o.g.f. x^2(2+x^3)/(1-x^3)^2. - Paul Barry, Jan 28 2004 [Artur Jasinski, Dec 11 2007, remarks that this should read (3*n + 2)*(1 + cos(2*Pi*(3*n + 2)/3 + Pi/3) - sqrt(3)*sin(2*Pi*(3*n + 2)/3 + Pi/3))/3.]
Except for 2, exponents e such that x^e + x + 1 is reducible. - N. J. A. Sloane, Jul 19 2005
The trajectory of these numbers under iteration of sum of cubes of digits eventually turns out to be 371 or 407 (47 is the first of the second kind). - Avik Roy (avik_3.1416(AT)yahoo.co.in), Jan 19 2009
Union of A165334 and A165335. - Reinhard Zumkeller, Sep 17 2009
a(n) is the set of numbers congruent to {2,5,8} mod 9. - Gary Detlefs, Mar 07 2010
It appears that a(n) is the set of all values of y such that y^3 = k*n + 2 for integer k. - Gary Detlefs, Mar 08 2010
These numbers do not occur in A000217 (triangular numbers). - Arkadiusz Wesolowski, Jan 08 2012
A089911(2*a(n)) = 9. - Reinhard Zumkeller, Jul 05 2013
Also indices of even Bell numbers (A000110). - Enrique Pérez Herrero, Sep 10 2013
Central terms of the triangle A108872. - Reinhard Zumkeller, Oct 01 2014
A092942(a(n)) = 1 for n > 0. - Reinhard Zumkeller, Dec 13 2014
a(n-1), n >= 1, is also the complex dimension of the manifold E(S), the set of all second-order irreducible Fuchsian differential equations defined on P^1 = C U {oo}, having singular points at most in S = {a_1, ..., a_n, a_{n+1} = oo}, a subset of P^1. See the Iwasaki et al. reference, Proposition 2.1.3., p. 149. - Wolfdieter Lang, Apr 22 2016
Except for 2, exponents for which 1 + x^(n-1) + x^n is reducible. - Ron Knott, Sep 16 2016
The reciprocal sum of 8 distinct items from this sequence can be made equal to 1, with these terms: 2, 5, 8, 14, 20, 35, 41, 1640. - Jinyuan Wang, Nov 16 2018
There are no positive integers x, y, z such that 1/a(x) = 1/a(y) + 1/a(z). - Jinyuan Wang, Dec 31 2018
As a set of positive integers, it is the set sum S + S where S is the set of numbers in A016777. - Michael Somos, May 27 2019
Interleaving of A016933 and A016969. - Leo Tavares, Nov 16 2021
Prepended with {1}, these are the denominators of the elements of the 3x+1 semigroup, the numerators being A005408 prepended with {2}. See Applegate and Lagarias link for more information. - Paolo Xausa, Nov 20 2021
This is also the maximum number of moves starting with n + 1 dots in the game of Sprouts. - Douglas Boffey, Aug 01 2022 [See the Wikipedia link. - Wolfdieter Lang, Sep 29 2022]
a(n-2) is the maximum sum of the span (or L(2,1)-labeling number) of a graph of order n and its complement. The extremal graphs are stars and their complements. For example, K_{1,2} has span 3, and K_2 has span 2. Thus a(3-1) = 5. - Allan Bickle, Apr 20 2023

			G.f. = 2 + 5*x + 8*x^2 + 11*x^3 + 14*x^4 + 17*x^5 + 20*x^6 + ... - _Michael Somos_, May 27 2019

K. Iwasaki, H. Kimura, S. Shimomura and M. Yoshida, From Gauss to Painlevé, Vieweg, 1991. p. 149.
Konrad Knopp, Theory and Application of Infinite Series, Dover, p. 269

G. C. Greubel, Table of n, a(n) for n = 0..10000
D. Applegate and J. C. Lagarias, The 3x+1 semigroup, Journal of Number Theory, Vol. 177, Issue 1, March 2006, pp. 146-159; see also the arXiv version, arXiv:math/0411140 [math.NT], 2004-2005.
H. Balakrishnan and N. Deo, Parallel algorithm for radiocoloring a graph, Congr. Numer. 160 (2003), 193-204.
Allan Bickle, Extremal Decompositions for Nordhaus-Gaddum Theorems, Discrete Math, 346 7 (2023), 113392.
L. Euler, Observatio de summis divisorum p. 9.
L. Euler, An observation on the sums of divisors, arXiv:math/0411587 [math.HO], 2004-2009, p. 9.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 937
L. B. W. Jolley, Summation of Series, Dover, 1961, p. 16
Tanya Khovanova, Recursive Sequences
Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original German edition of "Theory and Application of Infinite Series")
Fabian S. Reid, The Visual Pattern in the Collatz Conjecture and Proof of No Non-Trivial Cycles, arXiv:2105.07955 [math.GM], 2021.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.
Leo Tavares, Illustration: Capped Triangular Frames
Wikipedia, Sprouts (game)
Index entries for linear recurrences with constant coefficients, signature (2,-1).

First differences of A005449.
Cf. A002939, A017041, A017485, A125202, A017233, A179896, A017617, A016957, A008544 (partial products), A032766, A016777, A124388, A005351.
Cf. A087370.
Cf. similar sequences with closed form (2*k-1)*n+k listed in A269044.
Cf. A000096, A000217.
Cf. A016933, A016969.

List([0..70],n->3*n+2); # Muniru A Asiru, Nov 02 2018

a016789 = (+ 2) . (* 3)  -- Reinhard Zumkeller, Jul 05 2013

[3*n+2: n in [0..80]]; // Vincenzo Librandi, Apr 14 2015

seq(3*n+2, n = 0 .. 50); # Matt C. Anderson, May 18 2017

Range[2, 500, 3] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)
LinearRecurrence[{2,-1},{2,5},70] (* Harvey P. Dale, Aug 11 2021 *)

vector(100,n,3*n-1) \\ Derek Orr, Apr 13 2015

for n in range(0,100): print(3*n+2, end=', ') # Stefano Spezia, Nov 21 2018

G.f.: (2+x)/(1-x)^2.
a(n) = 3 + a(n-1).
a(n) = 1 + A016777(n).
a(n) = A124388(n)/9.
a(n) = A125199(n+1,1). - Reinhard Zumkeller, Nov 24 2006
Sum_{n>=1} (-1)^n/a(n) = (1/3)*(Pi/sqrt(3) - log(2)). - Benoit Cloitre, Apr 05 2002
1/2 - 1/5 + 1/8 - 1/11 + ... = (1/3)*(Pi/sqrt(3) - log 2). [Jolley] - Gary W. Adamson, Dec 16 2006
Sum_{n>=0} 1/(a(2*n)*a(2*n+1)) = (Pi/sqrt(3) - log 2)/9 = 0.12451569... (see A196548). [Jolley p. 48 eq (263)]
a(n) = 2*a(n-1) - a(n-2); a(0)=2, a(1)=5. - Philippe Deléham, Nov 03 2008
a(n) = 6*n - a(n-1) + 1 with a(0)=2. - Vincenzo Librandi, Aug 25 2010
Conjecture: a(n) = n XOR A005351(n+1) XOR A005352(n+1). - Gilian Breysens, Jul 21 2017
E.g.f.: (2 + 3*x)*exp(x). - G. C. Greubel, Nov 02 2018
a(n) = A005449(n+1) - A005449(n). - Jinyuan Wang, Feb 03 2019
a(n) = -A016777(-1-n) for all n in Z. - Michael Somos, May 27 2019
a(n) = A007310(n+1) + (1 - n mod 2). - Walt Rorie-Baety, Sep 13 2021
a(n) = A000096(n+1) - A000217(n-1). See Capped Triangular Frames illustration. - Leo Tavares, Oct 05 2021

A016957 a(n) = 6*n + 4.

Original entry on oeis.org

4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, 70, 76, 82, 88, 94, 100, 106, 112, 118, 124, 130, 136, 142, 148, 154, 160, 166, 172, 178, 184, 190, 196, 202, 208, 214, 220, 226, 232, 238, 244, 250, 256, 262, 268, 274, 280, 286, 292, 298, 304, 310, 316, 322, 328

Offset: 0

N. J. A. Sloane

Number of 2 X n binary matrices avoiding simultaneously the right-angled numbered polyomino patterns (ranpp) (00;1), (01,1) and (11;0). An occurrence of a ranpp (xy;z) in a matrix A=(a(i,j)) is a triple (a(i1,j1), a(i1,j2), a(i2,j1)) where i1 < i2, j1 < j2 and these elements are in the same relative order as those in the triple (x,y,z). In general, the number of m X n 0-1 matrices in question is given by (n+2)*2^(m-1) + 2*m*(n-1) - 2 for m > 1 and n > 1. - Sergey Kitaev, Nov 12 2004
If Y is a 4-subset of an n-set X then, for n >= 4, a(n-4) is the number of 3-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 08 2007
4th transversal numbers (or 4-transversal numbers): Numbers of the 4th column of positive numbers in the square array of nonnegative and polygonal numbers A139600. Also, numbers of the 4th column in the square array A057145. - Omar E. Pol, May 02 2008
a(n) is the maximum number such that there exists an edge coloring of the complete graph with a(n) vertices using n colors and every subgraph whose edges are of the same color (subgraph induced by edge color) is planar. - Srikanth K S, Dec 18 2010
Also numbers having two antecedents in the Collatz problem: 12*n+8 and 2*n+1 (respectively A017617(n) and A005408(n)). - Michel Lagneau, Dec 28 2012
a(n) = 6n+4 has three undirected edges e1 = (3n+2, 6n+4), e2 = (6n+4, 12n+8) and e3 = (2n+1, 6n+4) in the Collatz graph of A006370. - Heinz Ebert, Mar 16 2021
Conjecture: this sequence contains some but not all, even numbers with odd abundance A088827. They appear in this sequence at indices A186424(n) - 1. - John Tyler Rascoe, Jul 09 2022

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189. - From N. J. A. Sloane, Dec 01 2012

Heinz Ebert, A Graph Theoretical Approach to the Collatz Problem, arXiv:1905.07575 [math.GM], 2019-2020.
Tanya Khovanova, Recursive Sequences.
Sergey Kitaev, On multi-avoidance of right angled numbered polyomino patterns, Integers: Electronic Journal of Combinatorial Number Theory, Vol. 4 (2004), Article A21, 20pp.
Index to sequences related to polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A000217, A005408, A006370, A008588, A008615, A016921, A016789, A016933, A016945, A016958, A016969, A017617, A017329, A057145, A067412, A088827, A139600, A139606, A157176, A186424.

a016957 = (+ 4) . (* 6)  -- Reinhard Zumkeller, Jul 05 2013

seq(6*n+4, n = 0 .. 50) # Matt C. Anderson, Jun 09 2017

Range[4, 1000, 6] (* Vladimir Joseph Stephan Orlovsky, May 27 2011 *)

makelist(6*n+4, n, 0, 30); /* Martin Ettl, Nov 12 2012 */

a(n)=6*n+4 \\ Charles R Greathouse IV, Jul 10 2016

A008615(a(n)) = n+1. - Reinhard Zumkeller, Feb 27 2008
a(n) = A016789(n)*2. - Omar E. Pol, May 02 2008
A157176(a(n)) = A067412(n+1). - Reinhard Zumkeller, Feb 24 2009
a(n) = sqrt(A016958(n)). - Zerinvary Lajos, Jun 30 2009
a(n) = 2*(6*n+1) - a(n-1) (with a(0)=4). - Vincenzo Librandi, Nov 20 2010
a(n) = floor((sqrt(36*n^2 - 36*n + 1) + 6*n + 1)/2). - Srikanth K S, Dec 18 2010
From Colin Barker, Jan 30 2012: (Start)
G.f.: 2*(2+x)/(1-2*x+x^2).
a(n) = 2*a(n-1) - a(n-2). (End)
A089911(2*a(n)) = 9. - Reinhard Zumkeller, Jul 05 2013
a(n) = 3 * A005408(n) + 1. - Fred Daniel Kline, Oct 24 2015
a(n) = A057145(n+2,4). - R. J. Mathar, Jul 28 2016
a(4*n+2) = 4 * a(n). - Zhandos Mambetaliyev, Sep 22 2018
Sum_{n>=0} (-1)^n/a(n) = sqrt(3)*Pi/18 - log(2)/6. - Amiram Eldar, Dec 10 2021
E.g.f.: 2*exp(x)*(2 + 3*x). - Stefano Spezia, May 29 2024

A089911 a(n) = Fibonacci(n) mod 12.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1

Offset: 0

Casey Mongoven, Nov 14 2003

From Reinhard Zumkeller, Jul 05 2013: (Start)
Sequence has been applied by several composers to 12-tone equal temperament pitch structure. The complete Fibonacci mod 12 system (a set of 10 periodic sequences) exhausts all possible ordered dyads; that is, every possible combination of two pitches is found in these sets.
a(A008594(n)) = 0;
a(A227144(n)) = 1;
a(3*A047522(n)) = 2;
a(A017569(n)) = a(2*A016933(n)) = a(4*A016777(n)) = 3;
a(2*A017629(n)) = a(3*A017137(n)) = a(6*A004767(n)) = 4;
a(A227146(n)) = 5;
a(nonexistent) = 6;
a(2*A017581(n)) = 7;
a(2*A017557(n)) = a(4*A016813(n)) = 8;
a(A017617(n)) = a(2*A016957(n)) = a(4*A016789(n)) = 9;
a(3*A047621(n)) = 10;
a(2*A017653(n)) = 11. (End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..1199
Ron Knott, Fibonacci Numbers and the Golden Section
M. Renault, The Fibonacci Sequence Modulo M
A. P. Shah, Fibonacci Sequence Modulo m, Fibonacci Quarterly, Vol.6, No.2 (1968), 139-141.
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.
Index entries for linear recurrences with constant coefficients, signature (1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1).

Cf. A000045, A001175, A003893, A010881.

a089911 n = a089911_list !! n
a089911_list = 0 : 1 : zipWith (\u v -> (u + v) `mod` 12)
                       (tail a089911_list) a089911_list
-- Reinhard Zumkeller, Jul 01 2013

[Fibonacci(n) mod 12: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

with(combinat,fibonacci); A089911 := proc(n) fibonacci(n) mod 12; end;

Table[Mod[Fibonacci[n], 12], {n, 0, 100}] (* Vincenzo Librandi, Feb 04 2014 *)

a(n)=fibonacci(n)%12 \\ Charles R Greathouse IV, Feb 03 2014

Has period of 24, restricted period 12 and multiplier 5.

a(n) = (a(n-1) + a(n-2)) mod 12, a(0) = 0, a(1) = 1.

More terms from Ray Chandler, Nov 15 2003

A092259 Numbers that are congruent to {4, 8} mod 12.

Original entry on oeis.org

4, 8, 16, 20, 28, 32, 40, 44, 52, 56, 64, 68, 76, 80, 88, 92, 100, 104, 112, 116, 124, 128, 136, 140, 148, 152, 160, 164, 172, 176, 184, 188, 196, 200, 208, 212, 220, 224, 232, 236, 244, 248, 256, 260, 268, 272, 280, 284, 292, 296, 304, 308, 316, 320, 328, 332

Offset: 1

Giovanni Teofilatto, Feb 19 2004

Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001651, A017617, A017569, A069587, A092899, A145439, A157697.

Fourth row of A092260.

[n : n in [0..400] | n mod 12 in [4, 8]]; // Wesley Ivan Hurt, May 21 2016

A092259:=n->6*n-3-(-1)^n: seq(A092259(n), n=1..100); # Wesley Ivan Hurt, May 21 2016

Table[6n-3-(-1)^n, {n, 80}] (* Wesley Ivan Hurt, May 21 2016 *)
LinearRecurrence[{1,1,-1},{4,8,16},60] (* Harvey P. Dale, Oct 07 2021 *)

G.f.: 4*x*(1+x+x^2) / ( (1+x)*(x-1)^2 ).
a(n) = 4 * A001651(n).
Iff phi(n) = phi(3n/2), then n is in A069587. - Labos Elemer, Feb 25 2004
a(n) = 12*(n-1)-a(n-1) (with a(1)=4). - Vincenzo Librandi, Nov 16 2010
From Wesley Ivan Hurt, May 21 2016: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3) for n>3.
a(n) = 6n - 3 - (-1)^n.
a(2n) = A017617(n-1) for n>1, a(2n-1) = A017569(n-1) for n>1.
a(n) = -a(1-n), a(n) = A092899(n) + 1 for n>0. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi*sqrt(3)/36. - Amiram Eldar, Dec 30 2021
From Amiram Eldar, Nov 24 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 1/sqrt(2) + 1/sqrt(6) (A145439).
Product_{n>=1} (1 + (-1)^n/a(n)) = sqrt(2/3) (A157697). (End)

Edited and extended by Ray Chandler, Feb 21 2004

A184005 a(n) = n - 1 + ceiling(3*n^2/4); complement of A184004.

Original entry on oeis.org

1, 4, 9, 15, 23, 32, 43, 55, 69, 84, 101, 119, 139, 160, 183, 207, 233, 260, 289, 319, 351, 384, 419, 455, 493, 532, 573, 615, 659, 704, 751, 799, 849, 900, 953, 1007, 1063, 1120, 1179, 1239, 1301, 1364, 1429, 1495, 1563, 1632, 1703, 1775, 1849, 1924, 2001, 2079, 2159, 2240, 2323, 2407, 2493, 2580, 2669, 2759

Offset: 1

Clark Kimberling, Jan 08 2011

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A184004.

[(6*n^2 + 8*n - (-1)^n - 7)/8: n in [1..80]]; // Vincenzo Librandi, Feb 09 2011

a=4/3; b=0;
Table[n+Floor[(a*n+b)^(1/2)],{n,80}]
Table[n-1+Ceiling[(n*n-b)/a],{n,60}]
Table[n - 1 + Ceiling[3 n^2/4], {n, 60}] (* or *) CoefficientList[ Series[x (1 + 2 x + x^2 - x^3)/((1 + x) (1 - x)^3), {x, 0, 60}], x] (* or *) Table[Round[(6 n^2 + 8 n - 7)/8], {n, 60}] (* Michael De Vlieger, Mar 23 2016 *)
LinearRecurrence[{2,0,-2,1},{1,4,9,15},60] (* Harvey P. Dale, Sep 16 2016 *)

my(x='x+O('x^200)); Vec(x*(1+2*x+x^2-x^3)/((1+x)*(1-x)^3)) \\ Altug Alkan, Mar 23 2016

def A184005(n): return n+((m:=3*n**2)>>2)-(not m&3) # Chai Wah Wu, Oct 01 2024

a(n) = 2*a(n-1) - 2*a(n-3) + 1*a(n-4).
From Bruno Berselli, Jan 25 2011: (Start)
G.f.: x*(1 + 2*x + x^2 - x^3)/((1 + x)*(1 - x)^3).
a(n) = (6*n^2 + 8*n - (-1)^n - 7)/8. (End)
a(n) = round((6*n^2 + 8*n - 7)/8). - Bruno Berselli, Jan 25 2011
From Paul Curtz, Feb 09 2011: (Start)
a(n) - a(n-1)  = A007494(n).
a(n) - a(n-2)  = 3*n - 1 = A016789(n-1).
a(n) - a(n-4)  = 6*n - 8 = A016957(n-2).
a(n) - a(n-8)  = 12*n - 40 = A017617(n-4).
a(n) - a(n-16) = 24*n - 176 = 8*A016789(n-8).
a(n) - a(n-32) = 48*n - 736 = 16*A016789(n-16). (End)
a(n) = n^2 - floor((n-2)^2/4). - Bruno Berselli, Jan 17 2017
a(n) = A002061(n+2) - A002620(n+4). - Anton Zakharov, May 17 2017
E.g.f.: (1/8)*(8 + (6*x^2 + 14*x -7)*exp(x) - exp(-x)). - G. C. Greubel, Jul 22 2017

A332515 Numbers k such that phi(k) == 8 (mod 12), where phi is the Euler totient function (A000010).

Original entry on oeis.org

15, 16, 20, 24, 25, 30, 33, 44, 50, 51, 64, 66, 68, 69, 80, 87, 92, 96, 102, 116, 120, 123, 138, 141, 159, 164, 165, 174, 176, 177, 188, 200, 212, 213, 220, 236, 246, 249, 255, 256, 264, 267, 272, 275, 282, 284, 289, 300, 303, 318, 320, 321, 330, 332, 339, 340

Offset: 1

Amiram Eldar, Feb 14 2020

nonn

Dence and Pomerance showed that the asymptotic number of the terms below x is ~ c2 * x/sqrt(log(x)), where c2 = (sqrt(2*sqrt(3))/(3*Pi)) * c3^(-1/2) * (2*c3 - c4) = 0.3284176245..., c3 = Product_{primes p == 2 (mod 3)} (1 + 1/(p^2-1)), and c4 = Product_{primes p == 2 (mod 3)} (1 - 1/(p+1)^2).

			25 is a term since phi(25) = 20 == 8 (mod 12).

Amiram Eldar, Table of n, a(n) for n = 1..10000
Thomas Dence and Carl Pomerance, Euler's function in residue classes, in: K. Alladi, P. D. T. A. Elliott, A. Granville and G. Tenenbaum (eds.), Analytic and Elementary Number Theory, Developments in Mathematics, Vol. 1, Springer, Boston, MA, 1998, pp. 7-20, alternative link.

Cf. A000010, A017617, A175646, A332511, A332512, A332513, A332514, A332516.

[k:k in [1..350]| EulerPhi(k) mod 12 eq 8]; // Marius A. Burtea, Feb 14 2020

Select[Range[400], Mod[EulerPhi[#], 12] == 8 &]

A166138 Trisection A022998(3n+1).

Original entry on oeis.org

1, 8, 7, 20, 13, 32, 19, 44, 25, 56, 31, 68, 37, 80, 43, 92, 49, 104, 55, 116, 61, 128, 67, 140, 73, 152, 79, 164, 85, 176, 91, 188, 97, 200, 103, 212, 109, 224, 115, 236, 121, 248, 127, 260, 133, 272, 139, 284, 145, 296, 151, 308, 157, 320, 163, 332, 169, 344, 175, 356, 181, 368, 187, 380, 193, 392

Offset: 0

Paul Curtz, Oct 08 2009

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A165988, A166304.

LinearRecurrence[{0,2,0,-1},{1,8,7,20},70] (* Harvey P. Dale, Aug 15 2012 *)
Table[If[OddQ@ #, #, 2 #] &[3 n + 1], {n, 0, 65}] (* or *)
CoefficientList[Series[(1 + 8 x + 5 x^2 + 4 x^3)/((1 - x)^2 (1 + x)^2), {x, 0, 65}], x] (* Michael De Vlieger, Apr 27 2016 *)

a(2n) = 6n+1 = A016921(n).
a(2n+1) = 12n+8 = A017617(n).
a(n) = 2*a(n-2)-a(n-4) = (3n+1)*(3-(-1)^n)/2.
From G. C. Greubel, Apr 26 2016: (Start)
O.g.f.: (1 + 8*x + 5*x^2 + 4*x^3)/((1 - x)^2*(1 + x)^2).
E.g.f.: (1/2)*(-1 + 3*x + (3+9*x)*exp(2*x))*exp(-x). (End)

A347839 An array of the positive integers congruent to 2 modulo 3 (A016789), read by antidiagonals upwards, giving the present triangle.

Original entry on oeis.org

2, 5, 8, 11, 20, 32, 14, 44, 80, 128, 17, 56, 176, 320, 512, 23, 68, 224, 704, 1280, 2048, 26, 92, 272, 896, 2816, 5120, 8192, 29, 104, 368, 1088, 3584, 11264, 20480, 32768, 35, 116, 416, 1472, 4352, 14336, 45056, 81920, 131072, 38, 140, 464, 1664, 5888, 17408, 57344, 180224, 327680, 524288

Offset: 1

Wolfdieter Lang, Oct 21 2021

This array a = (a(k, n))_{k >= 1,n >= 0} is underlying array A of A347834. See the first formula. It has a simple recurrence for the rows k, given the first column a(k, 0) = A347838(k), which lists the positive integers congruent to {2, 5, 11} modulo 12.
In the array one can add the negative of the powers of 4 as row for k = 0, i.e., -A000302(n), for n >= 0.
All positive numbers congruent to 2 modulo 3 (A017617) appear once in this array. Proof from the array A of A347834 of the positive integers congruent to {1,3,5,7} modulo 8, and the present first formula: The members of column n = 0 give all the positive integers congruent to {2, 5, 11} modulo 12 once, and the members of columns n >= 1 give all the positive integers congruent to 8 modulo 12 (A017617) once. These members combined lead to the positive integers congruent to 2 modulo 3.

			The array a(k, n) begins:
k \ n  0   1   2    3    4     5      6      7       8       9       10 ...
---------------------------------------------------------------------------
1:     2   8  32  128  512  2048   8192  32768  131072  524288  2097152 ...
2:     5  20  80  320 1280  5120  20480  81920  327680 1310720  5242880 ...
3:    11  44 176  704 2816 11264  45056 180224  720896 2883584 11534336 ...
4:    14  56 224  896 3584 14336  57344 229376  917504 3670016 14680064 ...
5:    17  68 272 1088 4352 17408  69632 278528 1114112 4456448 17825792 ...
6:    23  92 368 1472 5888 23552  94208 376832 1507328 6029312 24117248 ...
7:    26 104 416 1664 6656 26624 106496 425984 1703936 6815744 27262976 ...
8:    29 116 464 1856 7424 29696 118784 475136 1900544 7602176 30408704 ...
9:    35 140 560 2240 8960 35840 143360 573440 2293760 9175040 36700160 ...
10:   38 152 608 2432 9728 38912 155648 622592 2490368 9961472 39845888 ...
...
----------------------------------------------------------------------------
The triangle t(n,k) begins:
k \ n  0   1   2    3    4     5     6      7      8      9 ...
---------------------------------------------------------------
1:     2
2:     5   8
3:    11  20  32
4:    14  44  80  128
5:    17  56 176  320  512
6:    23  68 224  704 1280  2048
7:    26  92 272  896 2816  5120  8192
8:    29 104 368 1088 3584 11264 20480  32768
9:    35 116 416 1472 4352 14336 45056  81920 131072
10:   38 140 464 1664 5888 17408 57344 180224 327680 524288
...
-----------------------------------------------------------------

Cf. A347834, A016789, A017617.
The rows k are given by -A000302 (for k=0), A004171, A003947(n+1), A002089, 2*A002042, ...
The columns n are given by 4^n*A347838 for n >= 0.

A := (n, k) -> 4^n*(3*(k + iquo(k, 3)) - 1):
for k from 1 to 10 do seq(A(n, k), n = 0..10) od;
# Alternatively:
gf  := n -> (4^n*((z*(z*(7*z + 3) + 3) - 1)))/((z - 1)^2*(1 + z + z^2)):
ser := n -> series(gf(n), z, 12):
col := (n, len) -> seq(coeff(ser(n), z, k), k = 1..len):
seq(print(col(n, 10)), n = 0..10); # Peter Luschny, Oct 26 2021

A[n_, k_] := 4^n (3(k + Quotient[k, 3]) - 1);
Table[A[n-k, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Nov 07 2021, from Maple code *)

Array a:
a(k, n) = (3*A(k, n) + 1)/2, with the array A from A347834, for k >= 1, and n >= 0.
a(k, n) = 4^n*A347838(k) = 4^n*(2 + 3*k + 3*floor((k + 1)/3)).
Recurrence for rows k: a(k, n) = 4*a(k, n-1), for n >= 1, with a(k, 0) = A347838(k).
O.g.f.: expansion in z gives the o.g.f.s for rows k, also for k = 0: -A000302; expansion in x gives the o.g.f.s for columns n.
G(z, x) = (-1 + 3*z + 3*z^2 + 7*z^3)/((1 - z)*(1 - z^3)*(1 - 4*x)).
Triangle t:
t(k, n) = a(k-n, n), for k >= 1, and n = 0, 1, ..., k-1.

cp's OEIS Frontend

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