A075677 -id:A075677 - cp's OEIS frontend

A160348 Minimal recursive sequence such that if a(n) > 0 then always a(n) > a((f(2n+1)-1)/2), where f is defined by f(2n+1) = (3n+2)/A006519(3n+2) for n>=1, that is f(m) = A075677(2*m-1) for odd m.

0, 2, 1, 6, 7, 5, 3, 11, 4, 13, 14, 10, 15, 52, 12, 50, 53, 9, 54, 59, 51, 62, 63, 49, 60, 65, 8, 68, 69, 58, 16, 75, 61, 56, 76, 48, 77, 80, 64, 84, 85, 67, 78, 88, 57, 44

Offset: 0

Vladimir Shevelev, May 10 2009; corrected May 13 2009, May 19 2009

If the (3x+1)-Collatz conjecture is true, then this sequence is a permutation of the nonnegative integers.

			a(0)=0. Let m=3. Then f(m)=5, f^2(m)=1. The corresponding numbers n=(m-1)/2 are 1,2,0. By the condition, a(1) > a(2) > a(0)=0. Therefore let a(2)=1, a(1)=2. Furthermore, consider m=7. Then f(m)=11, f^2(m)=17, f^3(m)=13, f^4(m)=5. The corresponding numbers n=(m-1)/2 are 3,5,8,6,2 and, by the condition, a(3) > a(5) > a(8) > a(6) > a(2)=1. Therefore set a(6)=3 (the minimal value which yet did not appear), a(8)=4, a(5)=5, a(3)=6, etc.

Cf. A006519, A075677, A159885, A159945, A160198, A122458, A259667.

Name edited by Michel Marcus, Feb 01 2021

A329480 a(n) = (1 - A075677(n))/6 if 6|(A075677(n)-1) or a(n) = (A075677(n) + 1)/6 if 6|(A075677(n)+1).

Original entry on oeis.org

0, 1, 0, 2, -1, 3, 1, 4, -2, 5, 0, 6, -3, 7, 2, 8, -4, 9, -1, 10, -5, 11, 3, 12, -6, 13, 1, 14, -7, 15, 4, 16, -8, 17, -2, 18, -9, 19, 5, 20, -10, 21, 0, 22, -11, 23, 6, 24, -12, 25, -3, 26, -13, 27, 7, 28, -14, 29, 2, 30, -15, 31, 8, 32, -16, 33, -4, 34, -17

Offset: 1

Fabian S. Reid, Jun 07 2020

sign

A fractal sequence.
This sequence is related to the Collatz Problem and can be illustrated on a logarithmic spiral to determine the odd numbers in the trajectory of a natural number of the form 6x+1 or 6x-1 simply by moving forward if the integer is positive, backward if the integer is negative, and continuing this forward-backward movement indefinitely.
When formatted as a table T with 4 columns, the third column T(n,3) is equal to the sequence. - Ruud H.G. van Tol, Oct 16 2023

			For n = 2, A075677(2) = 5, so a(2) = 1.
For n = 9, A075677(9) = 13, so a(9) = -2.
From _Ruud H.G. van Tol_, Oct 16 2023: (Start)
Array T begins:
 n|k_1|__2|__3|__4|
 1|  0   1   0   2
 2| -1   3   1   4
 3| -2   5   0   6
 4| -3   7   2   8
 5| -4   9  -1  10
 6| -5  11   3  12
... (End)

Ruud H.G. van Tol, Table of n, a(n) for n = 1..10000
Commons.Wikimedia Integer Spiral
Fabian S. Reid, The Visual Pattern in the Collatz Conjecture and Proof of No Non-Trivial Cycles, arXiv:2105.07955 [math.GM], 2021.

Cf. A002450, A016789, A075677, A096773, A349414.

nterms=100;Table[r=(c=3(2n-1)+1)/2^IntegerExponent[c,2];If[Mod[r,6]==1,(1-r)/6,(1+r)/6],{n, nterms}] (* Paolo Xausa, Nov 28 2021 *)

a(n) = my(x=3*n-1); x>>=valuation(x, 2); if(1==x%6, 1-x, 1+x)/6; \\ Ruud H.G. van Tol, Oct 16 2023

a(n) = (1 - A075677(n))/6 when 1 = A075677(n) mod 6, or
a(n) = (A075677(n) + 1)/6 when 5 = A075677(n) mod 6.
From Ruud H.G. van Tol, Oct 16 2023: (Start)
a(4*n-1) = a(n).
T(n,1) = 1-n; T(n,2) = 2*n-1 = n - T(n,1); T(n,3) = T(floor((n-1)/4) + 1, (n-1) mod 4 + 1) = a(n); T(n,4) = 2*n = T(n,2) + 1. (End)

A008908 a(n) = (1 + number of halving and tripling steps to reach 1 in the Collatz (3x+1) problem), or -1 if 1 is never reached.

Original entry on oeis.org

1, 2, 8, 3, 6, 9, 17, 4, 20, 7, 15, 10, 10, 18, 18, 5, 13, 21, 21, 8, 8, 16, 16, 11, 24, 11, 112, 19, 19, 19, 107, 6, 27, 14, 14, 22, 22, 22, 35, 9, 110, 9, 30, 17, 17, 17, 105, 12, 25, 25, 25, 12, 12, 113, 113, 20, 33, 20, 33, 20, 20, 108, 108, 7, 28, 28, 28, 15, 15, 15, 103

Offset: 1

N. J. A. Sloane, Bill Gosper

The number of steps (iterations of the map A006370) to reach 1 is given by A006577, this sequence counts 1 more. - M. F. Hasler, Nov 05 2017

When Collatz 3N+1 function is seen as an isometry over the dyadics, the halving step necessarily following each tripling is not counted, hence N -> N/2, if even, but N -> (3N+1)/2, if odd. Counting iterations of this map until reaching 1 leads to sequence A064433. - Michael Vielhaber (vielhaber(AT)gmail.com), Nov 18 2009

R. K. Guy, Unsolved Problems in Number Theory, E16.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
J. C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly, 92 (1985), 3-23.
Nitrxgen, Collatz Calculator
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006577, A006370, A006667, A075677.

a008908 = length . a070165_row
-- Reinhard Zumkeller, May 11 2013, Aug 30 2012, Jul 19 2011

a:= proc(n) option remember; 1+`if`(n=1, 0,
      a(`if`(n::even, n/2, 3*n+1)))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jan 29 2021

Table[Length[NestWhileList[If[EvenQ[ # ], #/2, 3 # + 1] &, i, # != 1 &]], {i, 75}]

a(n)=my(c=1); while(n>1, n=if(n%2, 3*n+1, n/2); c++); c \\ Charles R Greathouse IV, May 18 2015

def a(n):
    if n==1: return 1
    x=1
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        x+=1
        if n<2: break
    return x
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Apr 15 2017

a(n) = A006577(n) + 1.
a(n) = f(n,1) with f(n,x) = if n=1 then x else f(A006370(n),x+1).
a(A033496(n)) = A159999(A033496(n)). - Reinhard Zumkeller, May 04 2009
a(n) = A006666(n) + A078719(n).
a(n) = length of n-th row in A070165. - Reinhard Zumkeller, May 11 2013

More terms from Larry Reeves (larryr(AT)acm.org), Apr 27 2001
"Escape clause" added to definition by N. J. A. Sloane, Jun 06 2017
Edited by M. F. Hasler, Nov 05 2017

A139391 Next odd term in Collatz trajectory with starting value n.

Original entry on oeis.org

1, 1, 5, 1, 1, 3, 11, 1, 7, 5, 17, 3, 5, 7, 23, 1, 13, 9, 29, 5, 1, 11, 35, 3, 19, 13, 41, 7, 11, 15, 47, 1, 25, 17, 53, 9, 7, 19, 59, 5, 31, 21, 65, 11, 17, 23, 71, 3, 37, 25, 77, 13, 5, 27, 83, 7, 43, 29, 89, 15, 23, 31, 95, 1, 49, 33, 101, 17, 13, 35, 107, 9, 55, 37, 113, 19, 29

Offset: 1

Reinhard Zumkeller, Apr 17 2008

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Friedrich L. Bauer, Der (ungerade) Collatz-Baum, Informatik Spektrum 31 (Springer, April 2008).
Eric Weisstein's World of Mathematics, Collatz Problem.
Wikipedia, Collatz conjecture.
Index entries for sequences related to 3x+1 (or Collatz) problem.

Cf. A000265, A006370, A014682, A016825, A160967, A350091.

Cf. A075677 (odd bisection).

a[n_]:=Select[NestWhileList[If[EvenQ[#],#/2,3#+1] &,n,#>1 &],OddQ]; Prepend[Table[If[EvenQ[n],a[n][[1]],a[n][[2]]],{n,2,77}],1] (* Jayanta Basu, May 27 2013 *)

a(n) = my(x = if(n%2, 3*n+1, n/2)); x/2^valuation(x, 2); \\ Michel Marcus, Feb 27 2022

# first formula
def A006370(n): return 3*n+1 if n%2 else n//2
def a(n): return x if (x := A006370(n))%2 else a(x)
print([a(n) for n in range(1, 78)]) # Michael S. Branicky, Dec 15 2021

# fourth formula, uses A006370 above
def A000265(n):
    while n%2 == 0: n //= 2
    return n
def a(n): return A000265(A006370(n))
print([a(n) for n in range(1, 78)]) # Michael S. Branicky, Dec 15 2021

a(n) = A006370(n) if A006370(n) is odd, otherwise a(A006370(n)).
a(n) = A006370(n) iff n mod 4 = 2;
a(A016825(n)) = A006370(A016825(n));
a(n) = A000265(A006370(n)).
a(A160967(n)) = 1. - Reinhard Zumkeller, May 31 2009
For odd n, a(n) = a(2*A350091((n-1)/2)+1). - Ruud H.G. van Tol, Dec 17 2021
Sum_{k=1..n} a(k) ~ n^2 / 3. - Amiram Eldar, Aug 26 2024
a(n) = A000265(A014682(n)). - Alan Michael Gómez Calderón, Apr 10 2025

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

A371094 a(n) = m*(2^e) + ((4^e)-1)/3, where m = 3n+1, and e is the 2-adic valuation of m.

Original entry on oeis.org

1, 21, 7, 21, 13, 341, 19, 45, 25, 117, 31, 69, 37, 341, 43, 93, 49, 213, 55, 117, 61, 5461, 67, 141, 73, 309, 79, 165, 85, 725, 91, 189, 97, 405, 103, 213, 109, 1877, 115, 237, 121, 501, 127, 261, 133, 1109, 139, 285, 145, 597, 151, 309, 157, 5461, 163, 333, 169, 693, 175, 357, 181, 1493, 187, 381, 193, 789, 199

Offset: 0

Antti Karttunen (proposed by Ali Sada), Apr 19 2024

Construction: take the binary expansion of 3n+1 (A016777(n)), and substitute "01" for all trailing 0-bits that follow after its odd part (= A067745(1+n)), of which there are A371093(n) in total. See the examples.

			For n=1, 3*n+1 = 4, "100" in binary, when we substitute 01's for the two trailing 0's, we obtain 21, "10101" in binary, therefore a(1) = 21.
For n=6, 3*6+1 = 19, "10011" in binary, and there are no trailing 0's, and no changes, therefore a(6) = 19.
For n=7, 3*7+1 = 22, "10110" in binary, with one trailing 0, which when replaced with 01 gives us 45, "101101" in binary, therefore a(7) = 45.
For n=229, there are e=4 trailing bit expansions 0 -> 01,
  3n+1 = binary  101011  0 0 0 0
  a(n) = binary  101011 01010101

Antti Karttunen, Table of n, a(n) for n = 0..16384
Michael De Vlieger, Plot binary expansion of a(n) arranged with smallest bit at bottom and largest at top, n increasing from left to right for n = 0..3071. Black indicates 1, white indicates 0.
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for sequences related to binary expansion of n

Cf. A016777, A067745, A075677, A086893, A096773, A372289, A371093.
Cf. A016921, A372351 (even and odd bisection), A372290 (numbers occurring in the latter).
Cf. arrays A372282, A372283 and A371096, A371100, A371102.
Cf. also A302338.

Array[#2*(2^#3) + ((4^#3) - 1)/3 & @@ {#1, #2, IntegerExponent[#2, 2]} & @@ {#, 3 #1 + 1} &, 67, 0] (* Michael De Vlieger, Apr 19 2024 *)

A371094(n) = { my(m=1+3*n, e=valuation(m,2)); ((m*(2^e)) + (((4^e)-1)/3)); };

def A371094(n): return ((m:=3*n+1)<<(e:=(~m & m-1).bit_length()))+((1<<(e<<1))-1)//3 # Chai Wah Wu, Apr 28 2024

a(n) = A372289(A016777(n)).

a(2n) = A016777(2n) = A016921(n).

A075680 For odd numbers 2n-1, the minimum number of iterations of the reduced Collatz function R required to yield 1. The function R is defined as R(k) = (3k+1)/2^r, with r as large as possible.

Original entry on oeis.org

0, 2, 1, 5, 6, 4, 2, 5, 3, 6, 1, 4, 7, 41, 5, 39, 8, 3, 6, 11, 40, 9, 4, 38, 7, 7, 2, 41, 10, 10, 5, 39, 8, 8, 3, 37, 42, 3, 6, 11, 6, 40, 1, 9, 9, 33, 4, 38, 43, 7, 7, 31, 12, 36, 41, 24, 2, 10, 5, 10, 34, 15, 39, 15, 44, 8, 8, 13, 32, 13, 3, 37, 42, 42, 6, 3, 11, 30, 11, 18, 35, 6, 40, 23

Offset: 1

T. D. Noe, Sep 25 2002

See A075677 for the function R applied to the odd numbers once. The 3x+1 conjecture asserts that a(n) is a finite number for all n. The function R applied to the odd numbers shows the essential behavior of the 3x+1 iterations.

Bisection of A006667. - T. D. Noe, Jun 01 2006

			a(4) = 5 because 7 is the fourth odd number and 5 iterations are needed: R(R(R(R(R(7)))))=1.

T. D. Noe, Table of n, a(n) for n = 1..5000

Cf. A075677.
Cf. A075684 for the largest number attained during the iteration.
Cf. A000265.
Cf. A060445 which also counts intermediate even steps.
Cf. A173732.

a075680 n = snd $ until ((== 1) . fst)
            (\(x, i) -> (a000265 (3 * x + 1), i + 1)) (2 * n - 1, 0)
-- Reinhard Zumkeller, Jan 08 2014

nextOddK[n_] := Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; (* assumes odd n *) Table[m=n; cnt=0; If[n>1, While[m=nextOddK[m]; cnt++; m!=1]]; cnt, {n, 1, 200, 2}]

a(n)=my(s); n+=n-1; while(n>1, n+=n>>1+1; if(n%2==0, n>>=valuation(n,2)); s++); s \\ Charles R Greathouse IV, Dec 22 2021

sub a {
  my $v = 2 * shift() - 1;
  my $c = 0;
  until (1 == $v) {
    $v = 3 * $v + 1;
    $v /= 2 until ($v & 1);
    $c += 1;
  }
  return $c;
} # Ruud H.G. van Tol, Nov 16 2021

a(n) = a(A173732(n-1) + 1) + 1 for n >= 2. - Alan Michael Gómez Calderón, Apr 10 2025

A257852 Array A read by upward antidiagonals in which the entry A(n,k) in row n and column k is defined by A(n,k) = (2^n(6k - 3 - 2*(-1)^n) - 1)/3, n,k >= 1.

Original entry on oeis.org

3, 1, 7, 13, 9, 11, 5, 29, 17, 15, 53, 37, 45, 25, 19, 21, 117, 69, 61, 33, 23, 213, 149, 181, 101, 77, 41, 27, 85, 469, 277, 245, 133, 93, 49, 31, 853, 597, 725, 405, 309, 165, 109, 57, 35, 341, 1877, 1109, 981, 533, 373, 197, 125, 65, 39

Offset: 1

L. Edson Jeffery, Jul 12 2015

Sequence is a permutation of the odd natural numbers.

Let N_1 denote the set of odd natural numbers, and let |y|_2 denote 2-adic valuation of y. Define the map F : N_1 -> N_1 by F(x) = (3*x + 1)/2^|3*x+1|_2 (cf. A075677). Then row n of A is the set of all x in N_1 for which |3*x + 1|_2 = A371093(x) = n. Hence F(A(n,k)) = 6*k - 3 - 2*(-1)^n.

			From _Ruud H.G. van Tol_, Oct 17 2023, corrected and extended by _Antti Karttunen_, Apr 18 2024: (Start)
Array A begins:
n\k|   1|   2|   3|   4|   5|   6|   7|   8| ...
---+---------------------------------------------
1  |   3,   7,  11,  15,  19,  23,  27,  31, ...
2  |   1,   9,  17,  25,  33,  41,  49,  57, ...
3  |  13,  29,  45,  61,  77,  93, 109, 125, ...
4  |   5,  37,  69, 101, 133, 165, 197, 229, ...
5  |  53, 117, 181, 245, 309, 373, 437, 501, ...
6  |  21, 149, 277, 405, 533, 661, 789, 917, ...
... (End)

Cf. A006370, A075677, A096773 (after its initial 0, column 1 of this array).
Cf. A004767, A017077, A082285, A238477 (rows 1-4).
Cf. A371092, A371093 (column and row indices for odd numbers).
Cf. also arrays A371095, A371096, A371097, A371100, A371101, A371102, A371103.

(* Array: *)
Grid[Table[(2^n*(6*k - 3 - 2*(-1)^n) - 1)/3, {n, 10}, {k, 10}]]
(* Array antidiagonals flattened: *)
Flatten[Table[(2^(n - k + 1)*(6*k - 3 - 2*(-1)^(n - k + 1)) - 1)/ 3, {n, 10}, {k, n}]]

up_to = 105;
A257852sq(n,k) = ((2^n * (6*k - 3 - 2*(-1)^n) - 1)/3);
A257852list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A257852sq((a-(col-1)),col))); (v); };
v257852 = A257852list(up_to);
A257852(n) = v257852[n]; \\ Antti Karttunen, Apr 18 2024

From Ruud H.G. van Tol, Oct 17 2023: (Start)
A(n,k+1) = A(n,k) + 2^(n+1).
A(n+2,k) = A(n,k)*4 + 1.
A(1,k) = A004767(k-1).
A(2,k) = A017077(k-1).
A(3,k) = A082285(k-1).
A(4,k) = A238477(k). (End)
For all odd positive numbers n, A(A371093(n), A371092(n)) = n. - Antti Karttunen, Apr 24 2024

A122458 "Dropping time" of the reduced Collatz iteration starting with 2n+1.

Original entry on oeis.org

0, 2, 1, 4, 1, 3, 1, 4, 1, 2, 1, 3, 1, 37, 1, 35, 1, 2, 1, 5, 1, 3, 1, 34, 1, 2, 1, 3, 1, 4, 1, 34, 1, 2, 1, 32, 1, 3, 1, 5, 1, 2, 1, 3, 1, 28, 1, 5, 1, 2, 1, 26, 1, 3, 1, 19, 1, 2, 1, 3, 1, 5, 1, 9, 1, 2, 1, 4, 1, 3, 1, 4, 1, 2, 1, 3, 1, 25, 1, 13, 1, 2, 1, 18, 1, 3, 1, 5, 1, 2, 1, 3, 1, 4, 1, 8, 1, 2, 1, 5

Offset: 0

T. D. Noe, Sep 08 2006

nonn

We count only the 3x+1 steps of the usual Collatz iteration. We stop counting when the iteration produces a number less than the initial 2n+1. For a fixed dropping time k, let N(k)=A100982(k) and P(k)=2^(A020914(k)-1). There are exactly N(k) odd numbers less than P(k) with dropping time k. Moreover, the sequence is periodic: if d is one of the N(k) odd numbers, then k=a(d)=a(d+i*P(k)) for all i>=0. This periodicity makes it easy to compute the average dropping time of the reduced Collatz iteration: Sum_{k>0} k*N(k)/P(k) = 3.492651852186... (A122791).

			a(3)=4 because, starting with 7, the iteration produces 11,17,13,5 and the last term is less than 7.
n = 13: the fr trajectory for 2*13+1 = 27 is 41, 31, 47, 71, 107, 161, 121, 91, 137, 103, 155, 233, 175, 263, 395, 593, 445, 167, 251, 377, 283, 425, 319, 479, 719, 1079, 1619, 2429, 911, 1367, 2051, 3077, 577, 433, 325, 61, 23, 35, 53, 5, 1 with 41 terms (without 27), hence fr^[37] = 23 < 27  and  a(13) = 37. - _Wolfdieter Lang_, Feb 20 2019

Victor Klee and Stan Wagon, Old and New Unsolved Problems in Plane Geometry and Number Theory, Mathematical Association of America (1991) pp. 225-229, 308-309. [called on p. 225 stopping time for 2n+1 and the function C(2*n+1) = A075677(n+1), n >= 0. - Wolfdieter Lang, Feb 20 2019]

T. D. Noe, Table of n, a(n) for n = 0..10000

Cf. A000265, A060445, A075677 (one step of the reduced Collatz iteration), A075680.

Cf. A087113 (indices of 1's), A017077 (indices of 2's), A122791 (limit mean).

nextOddK[n_]:=Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; dt[n_]:=Module[{m=n, cnt=0}, If[n>1, While[m=nextOddK[m]; cnt++; m>n]]; cnt]; Table[dt[n],{n,1,301,2}]

a(n) is the least k for which fr^[k](n) < 2*n + 1, for n >= 1 and k >= 1, where fr(n) = A075677(n+1) = A000265(3*n+2). No k satisfies this for n = 0: a(0) := 0 by convention. The dropping time a(n) is finite, for n >= 1, if the Collatz conjecture is true. - Wolfdieter Lang, Feb 20 2019

a(1+i*8) = 2, for i>=0, because A100982(2) = 1 is odd, and A020914(2) = 4 gives P(2) = 2^(4-1) = 8. - Ruud H.G. van Tol, Dec 19 2021

A372443 The n-th iterate of 27 with Reduced Collatz-function R, which gives the odd part of 3n+1.

Original entry on oeis.org

27, 41, 31, 47, 71, 107, 161, 121, 91, 137, 103, 155, 233, 175, 263, 395, 593, 445, 167, 251, 377, 283, 425, 319, 479, 719, 1079, 1619, 2429, 911, 1367, 2051, 3077, 577, 433, 325, 61, 23, 35, 53, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Antti Karttunen, May 01 2024

nonn

Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A000265, A075677, A371093.
Column 14 of A372283, Row 13 of A256598 (but only up to the first 1).
Row 1 of A372560.
From term 47 to the first 1 same as A088593.
Sequences derived from this one or related to:
  A372444,
  A372445 column index of a(n) in array A257852,
  A372362 the 2-adic valuation of 1 + 3*a(n), equal to row index of a(n) in array A257852,
  A372447 binary lengths minus 1,
  A372446 a(n) xored with the term of A086893 having the same binary length,
  A372453 a(n) minus the term of A086893 having the same binary length.

R(n) = { n = 1+3*n; n>>valuation(n, 2); };
A372443(n) = { my(x=27); while(n, x=R(x); n--); (x); };

a(0) = 27; for n > 0, a(n) = R(a(n-1)), where R(n) = (3*n+1)/2^A371093(n) = A000265(3*n+1).

For n > 0, a(n) = R(A372444(n-1)) = A000265(1+3*A372444(n-1)).

cp's OEIS Frontend

A160348 Minimal recursive sequence such that if a(n) > 0 then always a(n) > a((f(2n+1)-1)/2), where f is defined by f(2n+1) = (3n+2)/A006519(3n+2) for n>=1, that is f(m) = A075677(2*m-1) for odd m.

Views

Author

Keywords

Comments

Examples

Crossrefs

Extensions

A329480 a(n) = (1 - A075677(n))/6 if 6|(A075677(n)-1) or a(n) = (A075677(n) + 1)/6 if 6|(A075677(n)+1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A008908 a(n) = (1 + number of halving and tripling steps to reach 1 in the Collatz (3x+1) problem), or -1 if 1 is never reached.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Python

Formula

Extensions

A139391 Next odd term in Collatz trajectory with starting value n.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Python

Python

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Formula

Extensions

A371094 a(n) = m*(2^e) + ((4^e)-1)/3, where m = 3n+1, and e is the 2-adic valuation of m.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula

A075680 For odd numbers 2n-1, the minimum number of iterations of the reduced Collatz function R required to yield 1. The function R is defined as R(k) = (3k+1)/2^r, with r as large as possible.

Views

A257852 Array A read by upward antidiagonals in which the entry A(n,k) in row n and column k is defined by A(n,k) = (2^n(6k - 3 - 2*(-1)^n) - 1)/3, n,k >= 1.