A347834 -id:A347834 - cp's OEIS frontend

A002450 a(n) = (4^n - 1)/3.

0, 1, 5, 21, 85, 341, 1365, 5461, 21845, 87381, 349525, 1398101, 5592405, 22369621, 89478485, 357913941, 1431655765, 5726623061, 22906492245, 91625968981, 366503875925, 1466015503701, 5864062014805, 23456248059221, 93824992236885, 375299968947541

Offset: 0

N. J. A. Sloane

For n > 0, a(n) is the degree (n-1) "numbral" power of 5 (see A048888 for the definition of numbral arithmetic). Example: a(3) = 21, since the numbral square of 5 is 5(*)5 = 101(*)101(base 2) = 101 OR 10100 = 10101(base 2) = 21, where the OR is taken bitwise. - John W. Layman, Dec 18 2001
a(n) is composite for all n > 2 and has factors x, (3*x + 2*(-1)^n) where x belongs to A001045. In binary the terms greater than 0 are 1, 101, 10101, 1010101, etc. - John McNamara, Jan 16 2002
Number of n X 2 binary arrays with path of adjacent 1's from upper left corner to right column. - R. H. Hardin, Mar 16 2002
The Collatz-function iteration started at a(n), for n >= 1, will end at 1 after 2*n+1 steps. - Labos Elemer, Sep 30 2002 [corrected by Wolfdieter Lang, Aug 16 2021]
Second binomial transform of A001045. - Paul Barry, Mar 28 2003
All members of sequence are also generalized octagonal numbers (A001082). - Matthew Vandermast, Apr 10 2003
Also sum of squares of divisors of 2^(n-1): a(n) = A001157(A000079(n-1)), for n > 0. - Paul Barry, Apr 11 2003
Binomial transform of A000244 (with leading zero). - Paul Barry, Apr 11 2003
Number of walks of length 2n between two vertices at distance 2 in the cycle graph C_6. For n = 2 we have for example 5 walks of length 4 from vertex A to C: ABABC, ABCBC, ABCDC, AFABC and AFEDC. - Herbert Kociemba, May 31 2004
Also number of walks of length 2n + 1 between two vertices at distance 3 in the cycle graph C_12. - Herbert Kociemba, Jul 05 2004
a(n+1) is the number of steps that are made when generating all n-step random walks that begin in a given point P on a two-dimensional square lattice. To make one step means to mark one vertex on the lattice (compare A080674). - Pawel P. Mazur (Pawel.Mazur(AT)pwr.wroc.pl), Mar 13 2005
a(n+1) is the sum of square divisors of 4^n. - Paul Barry, Oct 13 2005
a(n+1) is the decimal number generated by the binary bits in the n-th generation of the Rule 250 elementary cellular automaton. - Eric W. Weisstein, Apr 08 2006
a(k) = [M^k]2,1, where M is the 3 X 3 matrix defined as follows: M = [1, 1, 1; 1, 3, 1; 1, 1, 1]. - _Simone Severini, Jun 11 2006
a(n-1) / a(n) = percentage of wasted storage if a single image is stored as a pyramid with a each subsequent higher resolution layer containing four times as many pixels as the previous layer. n is the number of layers. - Victor Brodsky (victorbrodsky(AT)gmail.com), Jun 15 2006
k is in the sequence if and only if C(4k + 1, k) (A052203) is odd. - Paul Barry, Mar 26 2007
This sequence also gives the number of distinct 3-colorings of the odd cycle C(2*n - 1). - Keith Briggs, Jun 19 2007
All numbers of the form m*4^m + (4^m-1)/3 have the property that they are sums of two squares and also their indices are the sum of two squares. This follows from the identity m*4^m + (4^m-1)/3 = 4(4(..4(4m + 1) + 1) + 1) + 1 ..) + 1. - Artur Jasinski, Nov 12 2007
For n > 0, terms are the numbers that, in base 4, are repunits: 1_4, 11_4, 111_4, 1111_4, etc. - Artur Jasinski, Sep 30 2008
Let A be the Hessenberg matrix of order n, defined by: A[1, j] = 1, A[i, i] := 5, (i > 1), A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n) = charpoly(A,1). - Milan Janjic, Jan 27 2010
This is the sequence A(0, 1; 3, 4; 2) = A(0, 1; 4, 0; 1) of the family of sequences [a, b : c, d : k] considered by G. Detlefs, and treated as A(a, b; c, d; k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
6*a(n) + 1 is every second Mersenne number greater than or equal to M3, hence all Mersenne primes greater than M2 must be a 6*a(n) + 1 of this sequence. - Roderick MacPhee, Nov 01 2010
Smallest number having alternating bit sum n. Cf. A065359.
For n = 1, 2, ..., the last digit of a(n) is 1, 5, 1, 5, ... . - Washington Bomfim, Jan 21 2011
Rule 50 elementary cellular automaton generates this sequence. This sequence also appears in the second column of array in A173588. - Paul Muljadi, Jan 27 2011
Sequence found by reading the line from 0, in the direction 0, 5, ... and the line from 1, in the direction 1, 21, ..., in the square spiral whose edges are the Jacobsthal numbers A001045 and whose vertices are the numbers A000975. These parallel lines are two semi-diagonals in the spiral. - Omar E. Pol, Sep 10 2011
a(n), n >= 1, is also the inverse of 3, denoted by 3^(-1), Modd(2^(2*n - 1)). For Modd n see a comment on A203571. E.g., a(2) = 5, 3 * 5 = 15 == 1 (Modd 8), because floor(15/8) = 1 is odd and -15 == 1 (mod 8). For n = 1 note that 3 * 1 = 3 == 1 (Modd 2) because floor(3/2) = 1 and -3 == 1 (mod 2). The inverse of 3 taken Modd 2^(2*n) coincides with 3^(-1) (mod 2^(2*n)) given in A007583(n), n >= 1. - Wolfdieter Lang, Mar 12 2012
If an AVL tree has a leaf at depth n, then the tree can contain no more than a(n+1) nodes total. - Mike Rosulek, Nov 20 2012
Also, this is the Lucas sequence V(5, 4). - Bruno Berselli, Jan 10 2013
Also, for n > 0, a(n) is an odd number whose Collatz trajectory contains no odd number other than n and 1. - Jayanta Basu, Mar 24 2013
Sum_{n >= 1} 1/a(n) converges to (3*(log(4/3) - QPolyGamma[0, 1, 1/4]))/log(4) = 1.263293058100271... = A321873. - K. G. Stier, Jun 23 2014
Consider n spheres in R^n: the i-th one (i=1, ..., n) has radius r(i) = 2^(1-i) and the coordinates of its center are (0, 0, ..., 0, r(i), 0, ..., 0) where r(i) is in position i. The coordinates of the intersection point in the positive orthant of these spheres are (2/a(n), 4/a(n), 8/a(n), 16/a(n), ...). For example in R^2, circles centered at (1, 0) and (0, 1/2), and with radii 1 and 1/2, meet at (2/5, 4/5). - Jean M. Morales, May 19 2015
From Peter Bala, Oct 11 2015: (Start)
a(n) gives the values of m such that binomial(4*m + 1,m) is odd. Cf. A003714, A048716, A263132.
2*a(n) = A020988(n) gives the values of m such that binomial(4*m + 2, m) is odd.
4*a(n) = A080674(n) gives the values of m such that binomial(4*m + 4, m) is odd. (End)
Collatz Conjecture Corollary: Except for powers of 2, the Collatz iteration of any positive integer must eventually reach a(n) and hence terminate at 1. - Gregory L. Simay, May 09 2016
Number of active (ON, black) cells at stage 2^n - 1 of the two-dimensional cellular automaton defined by "Rule 598", based on the 5-celled von Neumann neighborhood. - Robert Price, May 16 2016
From Luca Mariot and Enrico Formenti, Sep 26 2016: (Start)
a(n) is also the number of coprime pairs of polynomials (f, g) over GF(2) where both f and g have degree n + 1 and nonzero constant term.
a(n) is also the number of pairs of one-dimensional binary cellular automata with linear and bipermutive local rule of neighborhood size n+1 giving rise to orthogonal Latin squares of order 2^m, where m is a multiple of n. (End)
Except for 0, 1 and 5, all terms are Brazilian repunits numbers in base 4, and so belong to A125134. For n >= 3, all these terms are composite because a(n) = {(2^n-1) * (2^n + 1)}/3 and either (2^n - 1) or (2^n + 1) is a multiple of 3. - Bernard Schott, Apr 29 2017
Given the 3 X 3 matrix A = [2, 1, 1; 1, 2, 1; 1, 1, 2] and the 3 X 3 unit matrix I_3, A^n = a(n)(A - I_3) + I_3. - Nicolas Patrois, Jul 05 2017
The binary expansion of a(n) (n >= 1) consists of n 1's alternating with n - 1 0's. Example: a(4) = 85 = 1010101_2. - Emeric Deutsch, Aug 30 2017
a(n) (n >= 1) is the viabin number of the integer partition [n, n - 1, n - 2, ..., 2, 1] (for the definition of viabin number see comment in A290253). Example: a(4) = 85 = 1010101_2; consequently, the southeast border of the Ferrers board of the corresponding integer partition is ENENENEN, where E = (1, 0), N = (0, 1); this leads to the integer partition [4, 3, 2, 1]. - Emeric Deutsch, Aug 30 2017
Numbers whose binary and Gray-code representations are both palindromes (i.e., intersection of A006995 and A281379). - Amiram Eldar, May 17 2021
Starting with n = 1 the sequence satisfies {a(n) mod 6} = repeat{1, 5, 3}. - Wolfdieter Lang, Jan 14 2022
Terms >= 5 are those q for which the multiplicative order of 2 mod q is floor(log_2(q)) + 2 (and which is 1 more than the smallest possible order for any q). - Tim Seuré, Mar 09 2024
The order of 2 modulo a(n) is 2*n for n >= 2. - Joerg Arndt, Mar 09 2024

			Apply Collatz iteration to 9: 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5 and hence 16, 8, 4, 2, 1.
Apply Collatz iteration to 27: 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5 and hence 16, 8, 4, 2, 1. [Corrected by _Sean A. Irvine_ at the suggestion of Stephen Cornelius, Mar 04 2024]
a(5) = (4^5 - 1)/3 = 341 = 11111_4 = {(2^5 - 1) * (2^5 + 1)}/3 = 31 * 33/3 = 31 * 11. - _Bernard Schott_, Apr 29 2017

A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 112.
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
Peter Bala, A characterization of A002450, A020988 and A080674.
Henry Bottomley, Illustration of initial terms
Sung-Hyuk Cha, On Complete and Size Balanced k-ary Tree Integer Sequences, International Journal of Applied Mathematics and Informatics, Issue 2, Volume 6, 2012, pp. 67-75. - From _N. J. A. Sloane_, Dec 24 2012
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. II. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See p. 8.
Carlos M. da Fonseca and Anthony G. Shannon, A formal operator involving Fermatian numbers, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
D. Dumont, Interprétations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318. (Annotated scanned copy)
David Eppstein, Making Change in 2048, arXiv:1804.07396 [cs.DM], 2018.
C. Ernst and D. W. Sumners, The Growth of the Number of Prime Knots, Math. Proc. Cambridge Philos. Soc. 102, 303-315, 1987
Ernesto Estrada and José A. de la Peña, Integer sequences from walks in graphs, Notes on Number Theory and Discrete Mathematics, Vol. 19, 2013, No. 3, 78-84.
Rigoberto Flórez, Robinson A. Higuita, and Antara Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Enrico Formenti and Luca Mariot, Exhaustive Generation of Linear Orthogonal CA, Automata 2023, Univ. Twente (Netherlands), see p. 22 of 38.
Mattia Fregola, Elementary Cellular Automata Rule 1 generating OEIS sequence A277799, A058896, A141725, A002450
A. Frosini and S. Rinaldi, On the Sequence A079500 and Its Combinatorial Interpretations, J. Integer Seq., Vol. 9 (2006), Article 06.3.1.
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 373
Petro Kosobutskyy and Nataliia Nestor, On the mathematical model of the transformation of natural numbers by a function of a split type, Comp. Des. Sys. Theor. Practice (2024) Vol. 6, No. 2. See p. 7.
Petro Kosobutskyy, Anastasiia Yedyharova, and Taras Slobodzyan, From Newton's binomial and Pascal's triangle to Collatz's problem, Comp. Des. Sys., Theor. Practice (2023) Vol. 5, No. 1, 121-127.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
J. V. Leyendekkers and A.G. Shannon, Modular Rings and the Integer 3, Notes on Number Theory & Discrete Mathematics, 17 (2011), 47-51.
Luca Manzoni, Luca Mariot, and Giuliamaria Menara, Combinatorial Designs and Cellular Automata: A Survey, arXiv:2503.10320 [math.CO], 2025. See pp. 16, 27.
Luca Mariot, Cryptography by Cellular Automata, 2017.
Luca Mariot, Orthogonal labelings in de Bruijn graphs, IWOCA 2020 - Open Problems Session, Delft University of Technology (Netherlands).
Luca Mariot, Connections between Latin squares, Cellular Automata and Coprime Polynomials, Univ. Twente (Netherlands, 2023). See p. 16/37.
Luca Mariot and Enrico Formenti, The number of coprime/non-coprime pairs of polynomials over F_2 with degree n and nonzero constant term.
Luca Mariot, Enrico Formenti and Alberto Leporati, Constructing Orthogonal Latin Squares from Linear Cellular Automata. In: Exploratory papers of AUTOMATA 2016.
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Index entries for linear recurrences with constant coefficients, signature (5,-4).

Partial sums of powers of 4, A000302.
When converted to binary, this gives A094028.
Subsequence of A003714.
Primitive factors: A129735.
Cf. A000225, A002446, A006995, A007583, A018215, A020988, A024036, A047849, A048716, A080355, A080674, A112627, A113860, A139391, A155701, A156605, A160967, A163834, A163868, A178415, A263132, A281379, A347834.

List([0..25], n -> (4^n-1)/3); # Muniru A Asiru, Feb 18 2018

a002450 = (`div` 3) . a024036
a002450_list = iterate ((+ 1) . (* 4)) 0
-- Reinhard Zumkeller, Oct 03 2012

[ (4^n-1)/3: n in [0..25] ]; // Klaus Brockhaus, Oct 28 2008

[n le 2 select n-1 else 5*Self(n-1)-4*Self(n-2): n in [1..70]]; // Vincenzo Librandi, Jun 13 2015

[seq((4^n-1)/3,n=0..40)];
A002450:=1/(4*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation, dropping the initial zero

Table[(4^n - 1)/3, {n, 0, 127}] (* Vladimir Joseph Stephan Orlovsky, Sep 29 2008 *)
LinearRecurrence[{5, -4}, {0, 1}, 30] (* Harvey P. Dale, Jun 23 2013 *)

makelist((4^n-1)/3, n, 0, 30); /* Martin Ettl, Nov 05 2012 */

PARI
```
a(n) = (4^n-1)/3;
```

my(z='z+O('z^40)); Vec(z/((1-z)*(1-4*z))) \\ Altug Alkan, Oct 11 2015

def A002450(n): return ((1<<(n<<1))-1)//3 # Chai Wah Wu, Jan 29 2023

((List.fill(20)(4: BigInt)).scanLeft(1: BigInt)( * )).scanLeft(0: BigInt)( + ) // Alonso del Arte, Sep 17 2019

From Wolfdieter Lang, Apr 24 2001: (Start)
a(n+1) = Sum_{m = 0..n} A060921(n, m).
G.f.: x/((1-x)*(1-4*x)). (End)
a(n) = Sum_{k = 0..n-1} 4^k; a(n) = A001045(2*n). - Paul Barry, Mar 17 2003
E.g.f.: (exp(4*x) - exp(x))/3. - Paul Barry, Mar 28 2003
a(n) = (A007583(n) - 1)/2. - N. J. A. Sloane, May 16 2003
a(n) = A000975(2*n)/2. - N. J. A. Sloane, Sep 13 2003
a(n) = A084160(n)/2. - N. J. A. Sloane, Sep 13 2003
a(n+1) = 4*a(n) + 1, with a(0) = 0. - Philippe Deléham, Feb 25 2004
a(n) = Sum_{i = 0..n-1} C(2*n - 1 - i, i)*2^i. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
a(n+1) = Sum_{k = 0..n} binomial(n+1, k+1)*3^k. - Paul Barry, Aug 20 2004
a(n) = center term in M^n * [1 0 0], where M is the 3 X 3 matrix [1 1 1 / 1 3 1 / 1 1 1]. M^n * [1 0 0] = [A007583(n-1) a(n) A007583(n-1)]. E.g., a(4) = 85 since M^4 * [1 0 0] = [43 85 43] = [A007583(3) a(4) A007583(3)]. - Gary W. Adamson, Dec 18 2004
a(n) = Sum_{k = 0..n, j = 0..n} C(n, j)*C(j, k)*A001045(j - k). - Paul Barry, Feb 15 2005
a(n) = Sum_{k = 0..n} C(n, k)*A001045(n-k)*2^k = Sum_{k = 0..n} C(n, k)*A001045(k)*2^(n-k). - Paul Barry, Apr 22 2005
a(n) = A125118(n, 3) for n > 2. - Reinhard Zumkeller, Nov 21 2006
a(n) = Sum_{k = 0..n} 2^(n - k)*A128908(n, k), n >= 1. - Philippe Deléham, Oct 19 2008
a(n) = Sum_{k = 0..n} A106566(n, k)*A100335(k). - Philippe Deléham, Oct 30 2008
If we define f(m, j, x) = Sum_{k = j..m} binomial(m, k)*stirling2(k, j)*x^(m - k) then a(n-1) = f(2*n, 4, -2), n >= 2. - Milan Janjic, Apr 26 2009
a(n) = A014551(n) * A001045(n). - R. J. Mathar, Jul 08 2009
a(n) = 4*a(n-1) + a(n-2) - 4*a(n-3) = 5*a(n-1) - 4*a(n-2), a(0) = 0, a(1) = 1, a(2) = 5. - Wolfdieter Lang, Oct 18 2010
a(0) = 0, a(n+1) = a(n) + 2^(2*n). - Washington Bomfim, Jan 21 2011
A036555(a(n)) = 2*n. - Reinhard Zumkeller, Jan 28 2011
a(n) = Sum_{k = 1..floor((n+2)/3)} C(2*n + 1, n + 2 - 3*k). - Mircea Merca, Jun 25 2011
a(n) = Sum_{i = 1..n} binomial(2*n + 1, 2*i)/3. - Wesley Ivan Hurt, Mar 14 2015
a(n+1) = 2^(2*n) + a(n), a(0) = 0. - Ben Paul Thurston, Dec 27 2015
a(k*n)/a(n) = 1 + 4^n + ... + 4^((k-1)*n). - Gregory L. Simay, Jun 09 2016
Dirichlet g.f.: (PolyLog(s, 4) - zeta(s))/3. - Ilya Gutkovskiy, Jun 26 2016
A000120(a(n)) = n. - André Dalwigk, Mar 26 2018
a(m) divides a(m*n), in particular: a(2*n) == 0 (mod 5), a(3*n) == 0 (mod 3*7), a(5*n) == 0 (mod 11*31), etc. - M. F. Hasler, Oct 19 2018
a(n) = 4^(n-1) + a(n-1). - Bob Selcoe, Jan 01 2020
a(n) = A178415(1, n) = A347834(1, n-1), arrays, for n >= 1.  - Wolfdieter Lang, Nov 29 2021
a(n) = A000225(2*n)/3. - John Keith, Jan 22 2022
a(n) = A080674(n) + 1 = A047849(n) - 1 = A163834(n) - 2 = A155701(n) - 3 = A163868(n) - 4 = A156605(n) - 7. - Ray Chandler, Jun 16 2023
From Peter Bala, Jul 23 2025: (Start)
The following are examples of telescoping products. Cf. A016153:
Product_{k = 1..2*n} 1 + 2^k/a(k+1) = a(n+1)/A007583(n) = (4^(n+1) - 1)/(2*4^n + 1).
Hence, Product_{k >= 1} 1 + 2^k/a(k+1) = 2.
Product_{k >= 1} 1 - 2^k/a(k+1) = 2/5, since 1 - 2^n/a(n+1) = b(n)/b(n-1), where b(n) = 2 - 3/(1 - 2^(n+1)).
Product_{k >= 1} 1 + (-2)^k/a(k+1) = 2/3, since 1 + (-2)^n/a(n+1) = c(n)/c(n-1), where c(n) = 2 - 1/(1 + (-2)^(n+1)).
Product_{k >= 1} 1 - (-2)^k/a(k+1) = 6/5, since 1 - (-2)^n/a(n+1) = d(n)/d(n-1), where d(n) = 2 - 1/(1 - (-2)^(n+1)). (End)

A075677 Reduced Collatz function R applied to the odd integers: a(n) = R(2n-1), where R(k) = (3k+1)/2^r, with r as large as possible.

Original entry on oeis.org

1, 5, 1, 11, 7, 17, 5, 23, 13, 29, 1, 35, 19, 41, 11, 47, 25, 53, 7, 59, 31, 65, 17, 71, 37, 77, 5, 83, 43, 89, 23, 95, 49, 101, 13, 107, 55, 113, 29, 119, 61, 125, 1, 131, 67, 137, 35, 143, 73, 149, 19, 155, 79, 161, 41, 167, 85, 173, 11, 179, 91, 185, 47, 191, 97, 197

Offset: 1

T. D. Noe, Sep 25 2002

The even-indexed terms a(2i+2) = 6i+5 = A016969(i), i >= 0 [Comment corrected by Bob Selcoe, Apr 06 2015]. The odd-indexed terms are the same as A067745. Note that this sequence is A016789 with all factors of 2 removed from each term. Also note that a(4i-1) = a(i). No multiple of 3 is in this sequence. See A075680 for the number of iterations of R required to yield 1.
From Bob Selcoe, Apr 06 2015: (Start)
All numbers in this sequence appear infinitely often.
From Eq. 1 and Eq. 2 in Formulas: Eq. 1 is used with 1/3 of the numbers in this sequence, Eq. 2 is used with 2/3 of the numbers.
(End)
Empirical: For arbitrary m, Sum_{n=2..A007583(m)} (a(n) - a(n-1)) = 0. - Fred Daniel Kline, Nov 23 2015
From Wolfdieter Lang, Dec 07 2021: (Start)
Only positive numbers congruent to 1 or 5 modulo 6 appear.
i) For the sequence entry with value A016921(m), for m >= 0, that is, a value from {1, 7, 13, ...}, the indices n are given by the row of array A178415(2*m+1, k), for k >= 1.
ii) For the sequence entry with value A007528(m), for m >= 1, that is, a value from {5, 11, 17, ...}, the indices n are given by the row of array A178415(2*m, k), for k >= 1.
See also the array A347834 with permuted row numbers and columns k >= 0. (End)

			a(11) = 1 because 21 is the 11th odd number and R(21) = 64/64 = 1.
From _Wolfdieter Lang_, Dec 07 2021: (Start)
i) 1 (mod 6) entry 1 = A016921(0) appears for n = A178415(1, k) = A347834(1, k-1) (the arrays), for k >= 1, that is, for {1, 5, 21, ..} = A002450.
ii) 5 (mod 6) entry 11 = A007528(2) appears for n = A178415(4, k) = A347835(3, k-1) (the arrays), for k >= 1, that is, for {7, 29, 117, ..} = A072261. (End)

Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section E16, pp. 330-336.
Victor Klee and Stan Wagon, Old and new unsolved problems in plane geometry and number theory, The Mathematical Association of America, 1991, p. 225, C(2n+1) = a(n+1), n >= 0.
Jeffrey C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010; see p. 57, also (90-9), p. 306.

T. D. Noe, Table of n, a(n) for n = 1..1000
Marc Chamberland, Una actualizacio del problema 3x + 1, Butl. Soc. Catalana Mat. (18), 19-45, 2003.
Jeffrey C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly, 92 (1985), 3-23.
Fabian S. Reid, The Visual Pattern in the Collatz Conjecture and Proof of No Non-Trivial Cycles, arXiv:2105.07955 [math.GM], 2021.
Eric Weisstein's World of Mathematics, Collatz Problem.
Index entries for sequences related to 3x+1 (or Collatz) problem.

Cf. A000265, A000302, A002450, A007528, A016789, A016921, A016969, A065677, A072197, A072261, A075680, A081294, A178415, A191669, A329480, A347834.
Cf. A006370, A014682 (for non-reduced Collatz maps), A087230 (A371093), A371094.
Odd bisection of A139391.
Even bisection of A067745, which is also the odd bisection of this sequence.
After the initial 1, the second leftmost column of A256598.
Row 2 of A372283.

a075677 = a000265 . subtract 2 . (* 6) -- Reinhard Zumkeller, Jan 08 2014

f:=proc(n) local t1;
if n=1 then RETURN(1) else
t1:=3*n+1;
while t1 mod 2 = 0 do t1:=t1/2; od;
RETURN(t1); fi;
end;
# N. J. A. Sloane, Jan 21 2011

nextOddK[n_] := Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; (* assumes odd n *) Table[nextOddK[n], {n, 1, 200, 2}]
v[x_] := IntegerExponent[x, 2]; f[x_] := (3*x + 1)/2^v[3*x + 1]; Table[f[2*n - 1], {n, 66}] (* L. Edson Jeffery, May 06 2015 *)

a(n)=n+=2*n-1;n>>valuation(n,2) \\ Charles R Greathouse IV, Jul 05 2013

from sympy import divisors
def a(n):
    return max(d for d in divisors(n) if d % 2)
print([a(6*n - 2) for n in range(1, 101)]) # Indranil Ghosh, Apr 15 2017, after formula by Reinhard Zumkeller

a(n) = A000265(6*n-2) = A000265(3*n-1). - Reinhard Zumkeller, Jan 08 2014
From Bob Selcoe, Apr 05 2015: (Start)
For all n>=1 and for every k, there exists j>=0 dependent upon n and k such that either:
Eq. 1: a(n) = (3n-1)/2^(2j+1) when k = ((4^(j+1)-1)/3) mod 2^(2j+3). Alternatively: a(n) = A016789(n-1)/A081294(j+1) when k = A002450(j+1) mod A081294(j+2). Example: n=51; k=101 == 5 mod 32, j=1. a(51) = 152/8 = 19.
  or
Eq. 2: a(n) = (3n-1)/4^j when k = (5*2^(2j+1) - 1)/3 mod 4^(j+1). Alternatively: a(n) = A016789(n-1)/A000302(j) when k = A072197(j) mod A000302(j+1). Example: n=91; k=181 == 53 mod 64, j=2. a(91) = 272/16 = 17.
(End) [Definition corrected by William S. Hilton, Jul 29 2017]
a(n) = a(n + g*2^r) - 6*g, n > -g*2^r. Examples: n=59; a(59)=11, r=5. g=-1: 11 = a(27) = 5 - (-1)*6; g=1: 11 = a(91) = 17 - 1*6; g=2: 11 = a(123) = 23 - 2*6; g=3: 11 = a(155) = 29 - 3*6; etc. - Bob Selcoe, Apr 06 2015
a(n) = a((1 + (3*n - 1)*4^(k-1))/3), k>=1 (cf. A191669). - L. Edson Jeffery, Oct 05 2015
a(n) = a(4n-1). - Bob Selcoe, Aug 03 2017
a(n) = A139391(2n-1). - Antti Karttunen, May 06 2024
Sum_{k=1..n} a(k) ~ n^2. - Amiram Eldar, Aug 26 2024
G.f.: Sum_{k>=1} ((3 + 2*(-1)^k)*x^(3*2^(k - 1) - (-2)^k/3 + 1/3) + (3 - 2*(-1)^k)*x^(2^(k - 1) - (-2)^k/3 + 1/3))/(x^(2^k) - 1)^2. - Miles Wilson, Oct 26 2024

A072197 a(n) = 4*a(n-1) + 1 with a(0) = 3.

Original entry on oeis.org

3, 13, 53, 213, 853, 3413, 13653, 54613, 218453, 873813, 3495253, 13981013, 55924053, 223696213, 894784853, 3579139413, 14316557653, 57266230613, 229064922453, 916259689813, 3665038759253, 14660155037013, 58640620148053, 234562480592213, 938249922368853, 3752999689475413

Offset: 0

N. Rathankar (rathankar(AT)yahoo.com), Jul 03 2002

Let A be the Hessenberg matrix of order n, defined by: A[1, j] = 1, A[i, i] := 2, (i > 1), A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n - 1) = (-1)^n*charpoly(A, -2). - Milan Janjic, Jan 26 2010
Numbers whose binary representation is 11 together with n times 01. For example, 213 = 11010101 (2). - Omar E. Pol, Nov 22 2012
The Collatz-function starting with a(n) will terminate at 1 after 2*n + 7 steps. This is because 3*a(n) + 1 = 5*2^(2n + 1), and the Collatz-function starting with 5 terminates at 1 after 5 additional steps.  So for example, a(2) = 53; Collatz sequence starting with 53 follows: 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 (11 steps). - Bob Selcoe, Apr 03 2015
a(n) is also the sum of the numerator and denominator of the binary fractions 0.1, 0.101, 0.10101, 0.1010101... Thus 0.1 = 1/2 with 1 + 2 = 3, 0.101 = 1/2 + 1/8 = 5/8 with 5 + 8 = 13; 0.10101 = 1/2 + 1/8 + 1/32 = 21/32 with 21 + 32 = 53. - J. M. Bergot, Sep 28 2016
a(n), for n >= 2, is also the smallest odd number congruent to 5 modulo 8 for which the modified reduced Collatz map given in A324036 has n consecutive extra steps compared to the reduced Collatz map given in A075677. - Nicolas Vaillant, Philippe Delarue, Wolfdieter Lang, May 09 2019

			a(1) = 13 because a(0) = 3 and 4 * 3 + 1 = 13.
a(2) = 53 because a(1) = 13 and 4 * 13 + 1 = 53.
a(3) = 213 because a(2) = 53 and 4 * 53 + 1 = 213.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Petro Kosobutskyy and Volodymyr Karkulovskyy, Recurrence and structuring of sequences of transformations 3n+ 1 as arguments for confirmation of the Collatz hypothesis, Comp. Des. Sys., Theor. Practice (2023) Vol. 5, No. 1, pp. 28-33. See p. 29.
Petro Kosobutskyy, Anastasiia Yedyharova, and Taras Slobodzyan, From Newton's binomial and Pascal's triangle to Collatz's problem, Comp. Des. Sys., Theor. Practice (2023) Vol. 5, No. 1, pp. 121-127.
Index entries for linear recurrences with constant coefficients, signature (5,-4).
Index entries for sequences related to 3x+1 (or Collatz) problem.

Cf. A000302, A002066 (first differences), A002450, A007583, A020988, A075677, A178415, A324036, A347834.

[(10*4^n-1)/3: n in [0..30]]; // Vincenzo Librandi, Oct 31 2011

A072197:=n->(10*4^n - 1)/3: seq(A072197(n), n=0..30); # Wesley Ivan Hurt, Sep 29 2016

Table[(10(4^n) - 1)/3, {n, 0, 19}] (* Alonso del Arte, Nov 22 2012 *)
NestList[4#+1&,3,30] (* Harvey P. Dale, Mar 09 2019 *)

a(n)=10*4^n\3 \\ Charles R Greathouse IV, Apr 06 2016

Vec((3-2*x)/((1-x)*(1-4*x)) + O(x^30)) \\ Colin Barker, Sep 28 2016

a(n) = (10*4^n - 1)/3 = 10*A002450(n) + 3. - Henry Bottomley, Dec 02 2002
a(n) = 5*a(n-1) - 4*a(n-2), n > 1. - Vincenzo Librandi, Oct 31 2011
a(n) = 2^(2*(n + 1)) - (2^(2*n + 1) + 1)/3 = A000302(n + 1) - A007583(n). - Vladimir Pletser, Apr 12 2014
a(n) = (5*2^(2*n + 1) - 1)/3. - Bob Selcoe, Apr 03 2015
G.f.: (3-2*x)/((1-x)*(1-4*x)). - Colin Barker, Sep 28 2016
a(n) = A020988(n) + A020988(n+1) + 1 = 2*(A002450(n) + A002450(n+1)) + 1. - Yosu Yurramendi, Jan 24 2017
a(n) = A002450(n+1) + 2^(2*n+1). - Adam Michael Bere, May 13 2021
a(n) = a(n-1) + 5*2^(2*n-1), for n >= 1, with a(0) = 3. - Wolfdieter Lang, Aug 16 2021
a(n) = A178415(2,n+1) = A347834(2,n), arrays, for n >= 0. - Wolfdieter Lang, Nov 29 2021
E.g.f.: exp(x)*(10*exp(3*x) - 1)/3. - Elmo R. Oliveira, Apr 02 2025

More terms from Henry Bottomley, Dec 02 2002

A178415 Array T(n,k) of odd Collatz preimages read by antidiagonals.

Original entry on oeis.org

1, 3, 5, 9, 13, 21, 7, 37, 53, 85, 17, 29, 149, 213, 341, 11, 69, 117, 597, 853, 1365, 25, 45, 277, 469, 2389, 3413, 5461, 15, 101, 181, 1109, 1877, 9557, 13653, 21845, 33, 61, 405, 725, 4437, 7509, 38229, 54613, 87381, 19, 133, 245, 1621, 2901, 17749, 30037

Offset: 1

T. D. Noe, May 28 2010

Every odd number occurs uniquely in this array. See A178414.

			Array T begins:
.    1    5   21    85   341   1365    5461   21845    87381   349525
.    3   13   53   213   853   3413   13653   54613   218453   873813
.    9   37  149   597  2389   9557   38229  152917   611669  2446677
.    7   29  117   469  1877   7509   30037  120149   480597  1922389
.   17   69  277  1109  4437  17749   70997  283989  1135957  4543829
.   11   45  181   725  2901  11605   46421  185685   742741  2970965
.   25  101  405  1621  6485  25941  103765  415061  1660245  6640981
.   15   61  245   981  3925  15701   62805  251221  1004885  4019541
.   33  133  533  2133  8533  34133  136533  546133  2184533  8738133
.   19   77  309  1237  4949  19797   79189  316757  1267029  5068117
- _L. Edson Jeffery_, Mar 11 2015
From _Bob Selcoe_, Apr 09 2015 (Start):
n=5, j=13: T(5,3) = 277 = (13*4^3 - 1)/3;
n=6, j=17: T(6,4) = 725 = (17*2^7 - 1)/3.
(End)

T. D. Noe, T(n,k) for n = 1..50, by antidiagonals
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

Rows of array: -A007583(k-1) (n=0), A002450 (n=1), A072197(k-1) (n=2), A206374(n=3), A072261 (n=4), A323824 (n=5), A072262 (n=6), A330246 (n=7), A072201 (n=8), ...
Columns of array: A022998(n-1)/2 (k=0), A178414 (k=1), ...
Cf. A347834 (permuted rows of the array).

t[n_,1] := t[n,1] = If[OddQ[n],4n-3,2n-1]; t[n_,k_] := t[n,k] = 4*t[n,k-1]+1; Flatten[Table[t[n-i+1,i], {n,20}, {i,n}]]

From Bob Selcoe, Apr 09 2015 (Start):
T(n,k) = 4*T(n,k-1) + 1.
T(n,k) = T(1,k) + 2^(2k+1)*(n-1)/2 when n is odd;
T(n,k) = T(2,k) + 4^k*(n-2)/2 when n >= 2 and n is even. So equivalently:
T(n,k) = T(n-2,k) + 2^(2k+1) when n is odd; and
T(n,k) = T(n-2,k) + 4^k when n is even.
Let j be the n-th positive odd number coprime with 3. Then:
T(n,k) = (j*4^k - 1)/3 when j == 1 (mod 3); and
T(n,k) = (j*2^(2k-1) - 1)/3 when j == 2 (mod 3).
(End)
From Wolfdieter Lang, Sep 18 2021: (Start)
T(n, k) = ((3*n - 1)*4^k - 2)/6 if n is even, and ((3*n - 2)*4^k - 1)/3 if n is odd, for n >= 1 and  k >= 1. Also for n = 0: -A007583(k-1), with A007583(-1) = 1/2, and for k = 0: A022998(n-1)/2, with A022998(-1) = -1.
O.g.f. for array T (with row n = 0 and column k = 0; z for rows and x for columns): G(z, x) = (1/(2*(1-x)*(1-4*x)*(1-z^2)^2)) * ((2*x-4)*z^3 + (3-5*x)*z^2 + 2*x*z + 3*x - 1). (End)

A072261 a(n) = 4*a(n-1) + 1, a(1)=7.

Original entry on oeis.org

7, 29, 117, 469, 1877, 7509, 30037, 120149, 480597, 1922389, 7689557, 30758229, 123032917, 492131669, 1968526677, 7874106709, 31496426837, 125985707349, 503942829397, 2015771317589, 8063085270357, 32252341081429, 129009364325717, 516037457302869

Offset: 1

N. Rathankar (rathankar(AT)yahoo.com), Jul 08 2002

These are the integers N which on application of the Collatz function yield the number 11. The Collatz function: if N is an odd number then (3N+1)/2^r yields a positive odd integer for some value of r (which in this case is 11).
These numbers reach 11 in Collatz function iteration after 2(n+1) steps and so end in 1 after exactly 2n+18 steps. - Lambert Klasen (lambert.klasen(AT)gmx.de), Nov 08 2004
Numbers whose binary representation is 111 together with n - 1 times 01. For example, a(4) = 469 = 111010101 (2). - Omar E. Pol, Nov 24 2012

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-4).
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A002450, A072197, A072257, A072258, A072259, A072260, A072262, A099730, A178415, A347834.

List([1..25], n-> (11*4^n -2)/6); # G. C. Greubel, Jan 14 2020

[(11*4^n -2)/6: n in [1..25]]; // G. C. Greubel, Jan 14 2020

seq(coeff(series(x*(7-6*x)/((1-x)*(1-4*x)), x, n+1), x, n), n = 1..25); # G. C. Greubel, Jan 14 2020

a[n_]:= 4a[n-1] +1; a[1]=7; Table[a[n], {n,25}]
NestList[4#+1&,7,30] (* or *) LinearRecurrence[{5,-4},{7,29},30] (* Harvey P. Dale, Sep 04 2023 *)

Vec(x*(7-6*x)/((1-x)*(1-4*x)) + O(x^25)) \\ Colin Barker, Oct 27 2019

[(11*4^n -2)/6 for n in (1..25)] # G. C. Greubel, Jan 14 2020

a(n) = (11*4^n - 2)/6 = 22*A002450(n-1) + 7. - Lambert Klasen (lambert.klasen(AT)gmx.de), Nov 08 2004
From Colin Barker, Oct 27 2019: (Start)
G.f.: x*(7 - 6*x) / ((1 - x)*(1 - 4*x)).
a(n) = 5*a(n-1) - 4*a(n-2) for n>2. (End)
E.g.f.: (-9 - 2*exp(x) + 11*exp(4*x))/6. - G. C. Greubel, Jan 14 2020
a(n) = a(n-1) + 11*2^(2*n-3), for n >= 2, with a(1) = 7. - Wolfdieter Lang, Aug 16 2021
a(n) = A178415(4, n) = A347834(3, n-1), arrays, for n >= 1. - Wolfdieter Lang, Nov 29 2021

Edited and extended by Robert G. Wilson v, Jul 17 2002

More terms from Colin Barker, Oct 27 2019

A206374 a(n) = (7*4^n - 1)/3.

Original entry on oeis.org

2, 9, 37, 149, 597, 2389, 9557, 38229, 152917, 611669, 2446677, 9786709, 39146837, 156587349, 626349397, 2505397589, 10021590357, 40086361429, 160345445717, 641381782869, 2565527131477, 10262108525909, 41048434103637, 164193736414549, 656774945658197

Offset: 0

Brad Clardy, Feb 07 2012

First bisection of A062092 and A081253, second bisection of A097163. - Bruno Berselli, Feb 12 2012

Except a(0)=2, this is the 3rd row of table A178415. - Michel Marcus, Apr 13 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Mike Warburton, Ulam-Warburton Automaton - Counting Cells with Quadratics, arXiv:1901.10565 [math.CO], 2019. See Table 1.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A002450, A006666, A072197; A002042 (first differences), A178415, A347834.

Magma
```
[(7*4^n-1)/3 : n in [0..30]];
```

Table[(7(4^n) - 1)/3, {n, 0, 24}] (* Alonso del Arte, Feb 11 2012 *)
CoefficientList[Series[(2-x)/(1-5*x+4*x^2),{x,0,30}],x] (* or *) LinearRecurrence[{5,-4},{2,9},30] (* Vincenzo Librandi, Mar 20 2012 *)

vector(20,n,(7*4^(n-1)-1)/3) \\ Derek Orr, Apr 12 2015

G.f.: (2-x)/(1-5*x+4*x^2). - Bruno Berselli, Feb 12 2012
a(n) = A083597(n)+1. - Bruno Berselli, Feb 12 2012
a(n) = 4*a(n-1)+1 for n>0, a(0)=2. - Bruno Berselli, Oct 22 2015
a(n) = 7*A002450(n) + 2. - Yosu Yurramendi, Jan 24 2017
A006666(a(n)) = 2*n+11 for n > 0. - Juan Miguel Barga Pérez, Jun 18 2020
a(n) = 5*a(n-1) - 4*a(n-2) for n >= 2. - Wesley Ivan Hurt, Jun 30 2020
a(n) = A178415(3, n) = A347834(4, n-1), arrays, for n >= 1.- Wolfdieter Lang, Nov 29 2021

A072262 a(n) = 4*a(n-1) + 1, a(1)=11.

Original entry on oeis.org

11, 45, 181, 725, 2901, 11605, 46421, 185685, 742741, 2970965, 11883861, 47535445, 190141781, 760567125, 3042268501, 12169074005, 48676296021, 194705184085, 778820736341, 3115282945365, 12461131781461, 49844527125845

Offset: 1

N. Rathankar (rathankar(AT)yahoo.com), Jul 08 2002

These are the integers N which on application of the Collatz function yield the number 17. The Collatz function: if N is an odd number then (3N+1)/2^r yields a positive odd integer for some value of r (which in this case is 17).

Numbers whose binary representation is 1011 together with n - 1 times 01. For example, a(4) = 725 = 1011010101 (2). - Omar E. Pol, Nov 24 2012

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-4).
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A072257, A072258, A072259, A072260, A072261, A099730, A178415, A347834.

List([1..30], n-> (17*4^n -2)/6); # G. C. Greubel, Jan 14 2020

[(17*4^n -2)/6: n in [1..30]]; // G. C. Greubel, Jan 14 2020

seq( (17*4^n -2)/6, n=1..30); # G. C. Greubel, Jan 14 2020

a[n_]:= 4a[n-1] +1; a[1]=11; Table[a[n], {n,25}]
NestList[4#+1&,11,30] (* or *) LinearRecurrence[{5,-4},{11,45},30] (* Harvey P. Dale, Dec 25 2014 *)

vector(30, n, (17*4^n -2)/6) \\ G. C. Greubel, Jan 14 2020

[(17*4^n -2)/6 for n in (1..30)] # G. C. Greubel, Jan 14 2020

From Bruno Berselli, Dec 16 2011: (Start)
G.f.: x*(11-10*x)/(1-5*x+4*x^2).
a(n) = (17*2^(2*n-1) - 1)/3.
Sum_{i=1..n} a(i) = (a(n+1) - n + 1)/3 - 4. (End)
a(n) = 34*A002450(n-1) + 11 . - Yosu Yurramendi, Jan 24 2017
E.g.f.: (-15 - 2*exp(x) + 17*exp(4*x))/6. - G. C. Greubel, Jan 14 2020
a(n) = A178415(6, n) = A347834(5, n-1), arrays, for n >= 1. - Wolfdieter Lang, Nov 29 2021

Edited and extended by Robert G. Wilson v, Jul 17 2002

A347840 A surjective map of the positive numbers congruent to 5 modulo 8 (A004770) to the positive numbers congruent to 1, 3, or 7 modulo 8 (A047529).

Original entry on oeis.org

1, 3, 1, 7, 9, 11, 3, 15, 17, 19, 1, 23, 25, 27, 7, 31, 33, 35, 9, 39, 41, 43, 11, 47, 49, 51, 3, 55, 57, 59, 15, 63, 65, 67, 17, 71, 73, 75, 19, 79, 81, 83, 1, 87, 89, 91, 23, 95, 97, 99, 25, 103, 105, 107, 27, 111, 113, 115, 7, 119, 121

Offset: 1

Wolfdieter Lang, Oct 30 2021

This map is obtained from the array A(k, m) given in A347834. There all positive numbers congruent to 5 modulo 8 (A004770) appear uniquely in the columns for m >= 1, and the m = 0 column gives all numbers congruent to {1, 3, 7} (mod 8) (A047529). The surjective map is f: A004770 -> A047529, with b(n) = A004770(n) -> f(b(n)) = a(n).
See also the array A178415 which has permuted rows.
This maps all entries of each row k of the array A(k, m), given in A347834, with columns m >= 1 to the entry A(k, 0) = A047529(k), for k >= 1. The numbers b(n) appear once in the array A for columns m >= 1. Column A(k, 1) = A347836(k) gives the numbers congruent to {5, 32, 29} (mod 32), and each entry for columns m >= 2 is congruent to 21 (mod 32).
The surjective map of the numbers b(n) = 5 + 8*(n-1) = A004770(n), for n >= 1, to A047529 with element a(n), is computed by switching to the companion array A347839 of A347834, with the simple recurrence, removing all factors of 4, and then going back to array A347834. See the formula below. Thanks to Antti Karttunen for motivating me to simplify the prescription, and to add in A347834 the hint for the induction proof that all 5 (mod 8) numbers appear once in the columns n >= 1.
This map f is of interest in the context of the Collatz 3*n+1 conjecture. The (modified) rooted tree with only odd labeled nodes has for each row k of the array A(k, m) (A347834) the same precursor (or (modified) Collatz map given in A075677(n+1), for 2*n+1). Therefore, all nodes with labels b(n) == 5 (mod 8) can be represented by a(n). This leads to a further restricted Collatz tree with only node labels congruent to {1, 3, 7} (mod 8) (A047529).
An even further restricted Collatz tree has only node labels congruent to 1 (mod 8) (A017077), as any positive integer can be written as m*2^(v+1)+2^v-1 or (m,v) where v is the number of trailing 1-bits in binary, and for v > 1 the next odd Collatz successor of (m,v) is (3*m+1,v-1). - Ruud H.G. van Tol, Sep 13 2023

			The sequence a(n) begins: (b(n) = A004770(n))
-------------------------------------------------------------------------
n:     1  2  3  4  5  6  7  8  9 10 11 12  13  14  15  16  17  18  19  20
b(n):  5 13 21 29 37 45 53 61 69 77 85 93 101 109 117 125 133 141 149 157
a(n):  1  3  1  7  9 11  3 15 17 19  1 23  25  27   7  31  35  35   9  39
-------------------------------------------------------------------------
n:     21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37 ...
b(n): 165 173 181 189 197 205 213 221 229 237 245 253 261 269 277 285 293 ...
a(n):  41  43  11  47  49  51   3  55  57  59  15  63  65  67  17  71  73 ...
-----------------------------------------------------------------------------
n = 6, b(6) = 45 = 13 + 32*1, case a), a(6) = 3 + 8*1 = 11.
n = 7, b(7) = 53 = 21 + 32*1, case b)1), first instance, L(7) = 0, a(7) = 3 + 8*0 = 3.
n = 31, b(31) = 245 = 117 + 128*1, case b)1), second instance, L(31) = 1, a(31) = 7 + 8*1 = 15.
n = 11, b(11) = 85 = 21 + 64*1, A065883(1 + 3*1) = 1, c(11) = 1, case b)2)i), a(11) = 85 = A347834(1, 3).
n = 19, b(19) = 149 = 21 + 64*2, A065883(1 + 3*2) = 7, c(19) = (7 - 1)/3 = 2, case b)2)ii), a(n) = 4*2 + 1 = 9.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A004770, A047529, A075677, A178415, A347834, A347836, A349414.

A347840[n_] := NestWhile[Quotient[#, 4] &, 2*n - 1, Mod[#, 8] == 5 &];
Array[A347840, 100] (* Paolo Xausa, Jun 25 2025 *)

a(n) = n=2*n-1; while(5==n%8, n>>=2); n; \\ Ruud H.G. van Tol, Sep 13 2023

a(n) = (2*n-1)>>(valuation(3*n-1,2)\2*2); \\ Ruud H.G. van Tol, Sep 20 2023

a(n) = (2*A065883((3*b(n)+1)/2) - 1)/3, with b(n) = A004770(n), for n >= 1.

a(n) = A385109(8*(n-1)+5). - Ralf Stephan, Jun 18 2025

cp's OEIS Frontend

A347836 a(n) = 8*(n + floor(n/3)) - 3; second column of A347834.

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A347837 a(n) = 32*(n + floor(n/3)) - 11; third column of A347834.

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A002450 a(n) = (4^n - 1)/3.

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A075677 Reduced Collatz function R applied to the odd integers: a(n) = R(2n-1), where R(k) = (3k+1)/2^r, with r as large as possible.

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A072197 a(n) = 4*a(n-1) + 1 with a(0) = 3.

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A178415 Array T(n,k) of odd Collatz preimages read by antidiagonals.

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