A198584 Odd numbers producing exactly 3 odd numbers in the Collatz (3x+1) iteration.

3, 13, 53, 113, 213, 227, 453, 853, 909, 1813, 3413, 3637, 7253, 7281, 13653, 14549, 14563, 29013, 29125, 54613, 58197, 58253, 116053, 116501, 218453, 232789, 233013, 464213, 466005, 466033, 873813, 931157, 932053, 932067, 1856853, 1864021, 1864133

Offset: 1

Author

T. D. Noe, Oct 28 2011

Keywords

• nonn
• easy

Comments

One of the odd numbers is always 1. So besides a(n), there is exactly one other odd number, A198585(n), which is a term in A002450.

Sequences A228871 and A228872 show that there are two sequences here: the odd numbers in order and out of order. - T. D. Noe, Sep 12 2013

Start with the numbers in A350053. If k is in sequence then so is 4*k+1. - Ralf Stephan, Jun 18 2025

Examples

			The Collatz iteration of 113 is 113, 340, 170, 85, 256, 128, 64, 32, 16, 8, 4, 2, 1, which shows that 113, 85, and 1 are the three odd terms.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..1009

Crossrefs

Cf. A062053 (numbers producing 3 odds in their Collatz iteration).

Cf. A092893 (least number producing n odd numbers).

Cf. A198586-A198593 (odd numbers producing 2-10 odd numbers).

Cf. A228871, A228872.

Programs

Mathematica

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; t = {}; Do[If[Length[Select[Collatz[n], OddQ]] == 3, AppendTo[t, n]], {n, 1, 10000, 2}]; t

Python

# get n-th term in sequence
def isqrt(n):
  i=0
  while(i*i<=n):
    i+=1
  return i-1
for n in range (200):
  s = isqrt(3*n)//3
  a = s*3
  b = (a*a)//3
  c = n-b
  d = 4*(n*3+a+(c4*s+1)+(c>5*s+1))+5
  e = isqrt(d)
  f = e-1-( (d-e*e) >> 1 )
  r = ((((8<André Hallqvist, Jul 25 2019

Python

# just prints the sequence
for a in range (5,100,1):
  for b in range(a-8+4*(a&1),0,-6):
    print(( ((1<André Hallqvist, Aug 14 2019

Formula

Numbers of the form (2^m*(2^n-1)/3-1)/3 where n == 2 (mod 6) if m is even and n == 4 (mod 5) if m is odd. - Charles R Greathouse IV, Sep 09 2022

a(n) = (16*2^floor(b(n)) - 2^(2*floor((b(n) - 1)/2) + 3*floor(b(n)) - 6*(floor(b(n)/2) - floor((floor(b(n))^2 + 20)/12) + n) - 2))/9 - 1/3 where b(n) = sqrt(3)*sqrt(4*n - 3). - Alan Michael Gómez Calderón, Feb 02 2025