A075486 -id:A075486 - cp's OEIS frontend

A075484 Length of iteration-list when Collatz-function(A006370) is iterated with initial value 5^n.

1, 6, 24, 109, 26, 124, 147, 139, 100, 92, 115, 337, 135, 277, 181, 261, 240, 219, 286, 322, 451, 337, 303, 432, 243, 540, 408, 444, 304, 464, 438, 554, 484, 582, 517, 677, 462, 617, 1002, 539, 655, 709, 714, 737, 623, 708, 868, 723, 707, 676, 642, 833, 776

Offset: 0

Labos Elemer, Sep 26 2002

nonn

			n=2: 5^n=25, list={25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10.5, 16, 8, 4, 2, 1}, a(2)=24.

T. D. Noe, Table of n, a(n) for n = 0..1000

Cf. A006370, A008908, A074472, A075485, A075486, A075487, A075488.

Table[Length[NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, 5^n, # != 1 &]], {n, 0, 52}] (* Michael De Vlieger, Feb 25 2017 *)

a(n) = A008908(5^n).

A179118 Number of Collatz steps to reach 1 starting with 2^n + 1.

Original entry on oeis.org

1, 7, 5, 19, 12, 26, 27, 121, 122, 35, 36, 156, 113, 52, 53, 98, 99, 100, 101, 102, 72, 166, 167, 168, 169, 170, 171, 247, 173, 187, 188, 251, 252, 178, 179, 317, 243, 195, 196, 153, 154, 155, 156, 400, 326, 495, 496, 161, 162, 331, 332, 408, 471, 410, 411, 337, 338, 339, 340, 553

Offset: 0

Mitch Harris, Jan 04 2011

There are many long runs of consecutive terms that increase by 1 (see second conjecture in A277109). For n < 40000, the longest run has 1030 terms starting from a(33237) = 244868 and ending with a(34266) = 245897. - Dmitry Kamenetsky, Sep 30 2016

			a(1)=7 because the trajectory of 2^1+1=3 is (3,10,5,16,8,4,2,1).

Dmitry Kamenetsky, Table of n, a(n) for n = 0..9999 (terms n = 1...1000 from Mitch Harris)
MathOverflow, Introduced on Mathoverflow by Henrik Rüping

Cf. A000051, A006577, A070976, A074472, A075486, A193688 (starting with 2^n-1), , A179118, A277109.

CollatzNext[n_] := If[Mod[n, 2] == 0, n/2, 3 n + 1]; CollatzPath[n_] := CollatzPath[n] = Module[{k = n, l = {}}, While[k != 1, k = CollatzNext[k]; l = Append[l, k]]; l]; Collatz[n_] := Length[CollatzPath[n]]; Table[Collatz[2^n+1],{n,1,50}]
f[n_] := Length@ NestWhileList[If[OddQ@ #, 3 # + 1, #/2] &, 2^n + 1, # > 1 &] - 1; Array[f, 60] (* Robert G. Wilson v, Jan 05 2011 *)
Array[-1 + Length@ NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, 2^# + 1, # > 1 &] &, 60, 0] (* Michael De Vlieger, Nov 25 2018 *)

nbsteps(n)= s=n; c=0; while(s>1, s=if(s%2, 3*s+1, s/2); c++); c;
a(n) = nbsteps(2^n+1); \\ Michel Marcus, Oct 28 2018

def steps(a):
  if a==1:     return 0
  elif a%2==0: return 1+steps(a//2)
  else:        return 1+steps(a*3+1)
for n in range(60):
  print(n, steps((1<

a(n) = A006577(2^n+1) = A006577(A000051(n)).

a(n) = A075486(n) - 1. - T. D. Noe, Jan 17 2013

a(0)=1 prepended by Alois P. Heinz, Dec 12 2018

A075485 Length of iteration list when Collatz-function is iterated with initial value 2^n - 1.

Original entry on oeis.org

1, 8, 17, 18, 107, 108, 47, 48, 62, 63, 157, 158, 159, 160, 130, 131, 225, 226, 178, 179, 304, 305, 474, 475, 445, 446, 385, 386, 449, 450, 451, 452, 528, 529, 530, 531, 532, 533, 534, 535, 536, 537, 587, 588, 589, 590, 591, 592, 593, 594, 595, 596, 853, 854

Offset: 1

Labos Elemer, Sep 26 2002

nonn

Somewhat surprisingly, these iterations take almost twice as long as the iterations for 2^n + 1. See A075486. - T. D. Noe, Jan 17 2013

			n=4, 2^n - 1 = 15, list = {15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1}, so a(4) = 18.

Cf. A006370, A008908, A074472, A075484, A075486, A075487, A075488.

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; Table[Length[Collatz[2^n - 1]], {n, 100}] (* T. D. Noe, Jan 17 2013 *)

a(n) = A008908(2^n-1).

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A075484 Length of iteration-list when Collatz-function(A006370) is iterated with initial value 5^n.

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A179118 Number of Collatz steps to reach 1 starting with 2^n + 1.

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A075485 Length of iteration list when Collatz-function is iterated with initial value 2^n - 1.

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