A072261 -id:A072261 - cp's OEIS frontend

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.

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

A347834 An array A of the positive odd numbers, read by antidiagonals upwards, giving the present triangle T.

Original entry on oeis.org

1, 3, 5, 7, 13, 21, 9, 29, 53, 85, 11, 37, 117, 213, 341, 15, 45, 149, 469, 853, 1365, 17, 61, 181, 597, 1877, 3413, 5461, 19, 69, 245, 725, 2389, 7509, 13653, 21845, 23, 77, 277, 981, 2901, 9557, 30037, 54613, 87381, 25, 93, 309, 1109, 3925, 11605, 38229, 120149, 218453, 349525

Offset: 1

Wolfdieter Lang, Sep 20 2021

For the definition of this array A see the formula section.
The rows of A appear in a draft by Immmo O. Kerner in eqs. (1) and (2) as so-called horizontal sequences (horizontale Folgen). Thanks to Dr. A. Eckert for sending me this paper.
This array with entry A(k, n) becomes equal to the array T with T(n, k) given in A178415 by using a permutation of the rows, and changing the offset: A(k, n) = T(pe(k), n+1), with pe(3*(L+1)) = 4*(L+1), pe(1+3*L) = 1 + 2*L, pe(2+3*L) = 2*(1 + 2*L), for L >= 0. This permutation appears in A265667.
A proper sub-array is A238475(n, k) = A(1 + 3*(k-1), n-1), for k >= 1 and n >= 1.
In the directed Collatz tree with nodes labeled with only positive odd numbers (see A256598 for the paths), here called CTodd, the level L = 0 (on the top) has the node with label 1 as root. Because 1 -> 1 there is an arrow (a 1-cycle or loop) at the root. The level L = 1 consists of the nodes with labels A(1, n), for n >= 1, and each node is connected to 1 by a downwards directed arrow. The next levels for L >= 2 are obtained using the successor rule (used also by Kerner): S(u) = (4*u - 1)/3 if u == 1 (mod 3), (2*u - 1)/3 if u == 5 (mod 3), and there is no successor S(u) = empty if u = 3 (mod 6), that is, this node is a leaf.
However, each node with label u on level L >= 1, except a leaf, has as successors at level L + 1 not only the node with S(u) but all the nodes with labels A(S(u), n), for n >= 0.
In this way each node (also the root) of this CTodd has in-degree 1 and infinite out-degree (for L >= 2 there are infinitely many infinite outgoing arrows). All nodes with label A(k, n) with n >= 1, have the same precursor as the node A(k,0) in this tree for each k >= 1.
Except for the loop (1-cycle) for the root 1 there are no cycles in this directed tree CTodd.
That each number N = 5 + 8*K, for K >= 0 appears in array A for some column n >= 1 uniquely can be proved, using the fact of strictly increasing rows and columns, by showing that the columns n = 1, 2, ..., c contain all positive integers congruent to 5 modulo 8 except those of the positive congruence class A(1, c+1) modulo 2^(2*c+3) by induction on c. [added Dec 05 2021]
Row index k for numbers congruent to 5 modulo 8: Each number N = 5 + 8*K, for K >= 0, from A004770 is a member of row k of the array A starting with element A(k, 0) = (2*A065883(2 + 3*N) - 1)/3. For this surjective map see A347840. [simplified Dec 05 2021]
The Collatz conjecture can be reduced to the conjecture that in this rooted and directed tree CTodd each positive odd number appears as a label once, that is, all entries of the array A appear.

			The array A(k, n) begins:
k\n  0   1   2    3    4     5      6      7       8       9       10 ...
-------------------------------------------------------------------------
1:   1   5  21   85  341  1365   5461  21845   87381  349525  1398101
2:   3  13  53  213  853  3413  13653  54613  218453  873813  3495253
3:   7  29 117  469 1877  7509  30037 120149  480597 1922389  7689557
4:   9  37 149  597 2389  9557  38229 152917  611669 2446677  9786709
5:  11  45 181  725 2901 11605  46421 185685  742741 2970965 11883861
6:  15  61 245  981 3925 15701  62805 251221 1004885 4019541 16078165
7:  17  69 277 1109 4437 17749  70997 283989 1135957 4543829 18175317
8:  19  77 309 1237 4949 19797  79189 316757 1267029 5068117 20272469
9:  23  93 373 1493 5973 23893  95573 382293 1529173 6116693 24466773
10: 25 101 405 1621 6485 25941 103765 415061 1660245 6640981 26563925
...
--------------------------------------------------------------------
The triangle T(k, n) begins:
k\n  0  1   2    3    4     5     6      7      8      9 ...
------------------------------------------------------------
1:   1
2:   3  5
3:   7 13  21
4:   9 29  53   85
5:  11 37 117  213  341
6:  15 45 149  469 853   1365
7:  17 61 181  597 1877  3413  5461
8:  19 69 245  725 2389  7509 13653  21845
9:  23 77 277  981 2901  9557 30037  54613  87381
10: 25 93 309 1109 3925 11605 38229 120149 218453 349525
...
-------------------------------------------------------------
Row index k of array A, for entries 5 (mod 8).
213 = 5 + 8*26. K = 28 is even, (3*231+1)/16 = 40, A065883(40) = 10, hence A(k, 0) = N' = (10-1)/3 = 3, and k = 2. Moreover, n = log_4((3*213 + 1)/(3*A(2,0) + 1)) = log_4(64) = 3. 213 = A(2, 3).
85 = 5 + 8*10. K = 10 is even, (3*85 + 1)/16 = 16, A065883(16) = 1, N' = (1-1)/3 = 0 is even, hence A(k, 0) = 4*0 + 1 = 1, k = 1. 85 = A(1, 3).
61 = 5 + 8*7, K = 7 is odd, k = (7+1)/2 + ceiling((7+1)/4) = 6, and n = log_4((3*61 + 1)/(3*A(6,0) + 1)) = 1. 61 = A(6, 1).
----------------------------------------------------------------------------

Paolo Xausa, Table of n, a(n) for n = 1..11325 (first 150 antidiagonals, flattened).
Immo O. Kerner, Elementarer Lösungsansatz zum Collatz-Ulam-Problem CUP oder das "3a + 1" Problem, Version Nov 26 2000.
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

Row sequences of the array A, also diagonal sequences of the triangle T: -A007583 (k=0), A002450(n+1), A072197, A072261(n+1), A206374(n+1), A072262(n+1), A072262(n+1), A072201(n+1), A330246(n+1), ...
Column sequences of the array A, also of the triangle T (shifted): A047529, A347836, A347837, ...
Cf. A004770, A007494, A016777, A065883, A047395, A047529, A178415, A238475, A256598, A265667, A319451, A347839, A347840.

# Seen as an array:
A := (n, k) -> ((3*(n + floor(n/3)) - 1)*4^(k+1) - 2)/6:
for n from 1 to 6 do seq(A(n, k), k = 0..9) od;
# Seen as a triangle:
T := (n, k) -> 2^(2*k + 1)*(floor((n - k)/3) - k + n - 1/3) - 1/3:
for n from 1 to 9 do seq(T(n, k), k = 0..n-1) od;
# Using row expansion:
gf_row := k -> (1 / (x - 1) - A047395(k)) / (4*x - 1):
for k from 1 to 10 do seq(coeff(series(gf_row(k), x, 11), x, n), n = 0..10) od;
# Peter Luschny, Oct 09 2021

A347834[k_, n_] := (4^n*(6*(Floor[k/3] + k) - 2) - 1)/3;
Table[A347834[k - n, n], {k, 10}, {n, 0, k - 1}] (* Paolo Xausa, Jun 26 2025 *)

Array A:
A(k, 0) = A047529(k) (the positive odd numbers {1, 3, 7} (mod 8));
A(k, n) = ((3* A(k, 0) + 1)*4^n - 1)/3, for k >= 1 and n >= 0.
Recurrence for rows k >= 1: A(k, n) = 4*A(k, n-1) + 1, for n >= 1, with A(k, 0) = 2*(k + floor(k/3)) - 1 = A047529(k).
Explicit form: A(k, n) = ((3*(k + floor(k/3)) - 1)*4^(n+1) - 2)/6, k >= 1, n >= 0. Here 3*(k + floor(k/3)) = A319451(k).
Hence A(k, n) = 5 + 8*(2*A(k, n-2)), for n >= 1, with A(k, 0) = 2*(k + floor(k/3)) - 1 = A047529(k), and 2*A(k, -1) = (A(k, 1) - 5)/8 = k - 1 + floor(k/3) (equals index n of A(k, 1) in the sequence (A004770(n+1))_{n >= 0}). A(k, -1) is half-integer if k = A007494(m) = m + ceiling(m/2), for m >= 1, and A(k, -1) = 2*K if k = 1 + 3*K = A016777(K), for K >= 0.
O.g.f.: expansion in z gives o.g.f.s for rows k, also for k = 0: -A007583; expansion in x gives o.g.f.s for columns n.
  G(z, x) = (2*(-1 + 3*z + 3*z^2 + 7*z^3)*(1-x) - (1-4*x)*(1-z^3)) / (3*(1-x)*(1-4*x)*(1-z)*(1-z^3)).
Triangle T:
T(k, n) = A(k - n, n), for k >= 1 and n = 0..k-1.
A(k, n) = [x^n] (1/(x - 1) - A047395(k)) / (4*x - 1). - Peter Luschny, Oct 09 2021

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)

A002089 a(n) = 11*4^n.

Original entry on oeis.org

11, 44, 176, 704, 2816, 11264, 45056, 180224, 720896, 2883584, 11534336, 46137344, 184549376, 738197504, 2952790016, 11811160064, 47244640256, 188978561024, 755914244096, 3023656976384, 12094627905536, 48378511622144, 193514046488576, 774056185954304, 3096224743817216

Offset: 0

N. J. A. Sloane

Subsequence of A000069, the odious numbers. - Reinhard Zumkeller, Aug 26 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (4).

Bisection of A005015. Half the first differences of A072261.

[11*4^n: n in [0..30]]; // Vincenzo Librandi, May 31 2011

A002089:=n->11*4^n; seq(A002089(n), n=0..100); # Wesley Ivan Hurt, Nov 12 2013

11*4^Range[0, 100] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
NestList[4#&,11,30] (* Harvey P. Dale, Jun 19 2014 *)

a(n)=11*4^n \\ Charles R Greathouse IV, Apr 18 2012

From Philippe Deléham, Nov 23 2008: (Start)
a(n) = 4*a(n-1); a(0)=11.
G.f.: 11/(1-4*x). (End)
a(n) = 11*A000302(n). - Michel Marcus, Nov 13 2013
From Elmo R. Oliveira, Apr 02 2025: (Start)
E.g.f.: 11*exp(4*x).
a(n) = A005015(2*n). (End)

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

A178414 Least odd number in the Collatz (3x+1) preimage of odd numbers not a multiple of 3.

Original entry on oeis.org

1, 3, 9, 7, 17, 11, 25, 15, 33, 19, 41, 23, 49, 27, 57, 31, 65, 35, 73, 39, 81, 43, 89, 47, 97, 51, 105, 55, 113, 59, 121, 63, 129, 67, 137, 71, 145, 75, 153, 79, 161, 83, 169, 87, 177, 91, 185, 95, 193, 99, 201, 103, 209, 107, 217, 111, 225, 115, 233, 119, 241, 123, 249

Offset: 1

T. D. Noe, May 28 2010

The odd non-multiples of 3 are 1, 5, 7, 11,... (A007310). The odd multiples of 3 have no odd numbers their Collatz pre-image. The next odd number in the Collatz iteration of a(2n) is 6n-1. The next odd number in the Collatz iteration of a(2n+1) is 6n+1. For each non-multiple of 3, there are an infinite number of odd numbers in its Collatz pre-image. For example:
Odd pre-images of 1: 1, 5, 21, 85, 341,... (A002450)
Odd pre-images of 5: 3, 13, 53, 213, 853,... (A072197)
Odd pre-images of 7: 9, 37, 149, 597, 2389,...
Odd pre-images of 11: 7, 29, 117, 469, 1877,...(A072261)
In each case, the pre-image sequence is t(k+1) = 4*t(k) + 1 with t(0)=a(n). The array of pre-images is in A178415.
a(n) = A047529(P(n)), with the permutation P(n) = A006368(n-1) + 1, for n >= 1. This shows that this sequence gives the numbers {1, 3, 7} (mod 8) uniquely. - Wolfdieter Lang, Sep 21 2021

Matthew House, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A002450, A006368, A007310, A022998, A047529, A072197, A072261, A114752.

Riffle[1+8*Range[0,50], 3+4*Range[0,50]]

a(n) = (n - 1)*(3 - (-1)^n) + 1. [Bogart B. Strauss, Sep 20 2013, adapted to the offset by Matthew House, Feb 14 2017]
From Matthew House, Feb 14 2017: (Start)
G.f.: x*(1 + 3*x + 7*x^2 + x^3)/((1 - x)^2*(1 + x)^2).
a(n) = 2*a(n-2) - a(n-4). (End)
From Philippe Deléham, Nov 06 2023: (Start)
a(2*n) = 4*n-1, a(2*n+1) = 8*n+1.
a(n) = 2*A022998(n-1)+1.
a(n) = 2*A114752(n)-1. (End)

A099730 Array read by antidiagonals. Rows contain odd numbers reaching same odd successor in Collatz function iteration.

Original entry on oeis.org

1, 5, 3, 21, 13, 7, 85, 53, 29, 9, 341, 213, 117, 37, 11, 1365, 853, 469, 149, 45, 15, 5461, 3413, 1877, 597, 181, 61, 17, 21845, 13653, 7509, 2389, 725, 245, 69, 19, 87381, 54613, 30037, 9557, 2901, 981, 277, 77, 23

Offset: 1

Lambert Klasen (lambert.klasen(AT)gmx.de), Nov 09 2004

All numbers that end in 3 will begin with numbers from previous row (for example, 3413 is 341&3). - Jean-Bernard François, Sep 09 2013

The sequence is a permutation of the odd positive integers. - Bob Selcoe, Jul 26 2015

			t(1, 2) = 53 = 4*13+1, t(2, 5) = 7509 = 4*1877+1.
Array begins:
1 5 21 85 341 1365 5461 21845 87381 ...
3 13 53 213 853 3413 13653 54613 218453 ...
7 29 117 469 1877 7509 30037 120149 480597 ...
9 37 149 597 2389 9557 38229 152917 611669 ...
11 45 181 725 2901 11605 46421 185685 742741 ...
15 61 245 981 3925 15701 62805 251221 1004885 ...
17 69 277 1109 4437 17749 70997 283989 1135957 ...
19 77 309 1237 4949 19797 79189 316757 1267029 ...
...
Construct array by writing odd numbers in columns, taking first overflow after two steps and then an overflow each fourth step (for each column).

First row = A002450 (except leading zero), second row = A072197, third row = A072261.

t[n_, k_] := 2^(2*(k + 1) - 1)*(n + Quotient[n + 1, 3]) + (4^(k + 1) - 1)/3; Table[t[n - k, k], {n, 0, 8}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Sep 09 2013, after Maon Wenders *)

g(n)=(n+1)\3
T(n,k)=2^(2*(k+1)-1)*(n+g(n))+(4^(k+1)-1)/3
for(i=0,20,for(j=0,10,print1(T(i,j), ", "));print())\\ Maon Wenders, Jul 15 2012

Let g(n)= floor((n+1)/3), then T(n,k) = 2^(2*(k+1)-1) *(n+g(n)) + (4^(k+1)-1)/3. - Maon Wenders, Jul 15 2012

t(n, k) = 4*t(n, k-1) + 1. - Jean-Bernard François, Sep 09 2013

cp's OEIS Frontend

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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A347834 An array A of the positive odd numbers, read by antidiagonals upwards, giving the present triangle T.

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

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A002089 a(n) = 11*4^n.

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

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A178414 Least odd number in the Collatz (3x+1) preimage of odd numbers not a multiple of 3.

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A099730 Array read by antidiagonals. Rows contain odd numbers reaching same odd successor in Collatz function iteration.