A198585 -id:A198585 - cp's OEIS frontend

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

T. D. Noe, Oct 28 2011

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

			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.

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

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.

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

# 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

# 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

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

A198586 a(n) = (4^A001651(n+1) - 1)/3: numbers (4^k-1)/3 for k > 1, not multiples of 3.

Original entry on oeis.org

5, 85, 341, 5461, 21845, 349525, 1398101, 22369621, 89478485, 1431655765, 5726623061, 91625968981, 366503875925, 5864062014805, 23456248059221, 375299968947541, 1501199875790165, 24019198012642645, 96076792050570581, 1537228672809129301

Offset: 1

T. D. Noe, Oct 30 2011

Numbers coprime to 6 producing 2 odd numbers in the Collatz iteration.
Numbers appearing in A198585 (sorted and duplicates removed). These numbers occur in A002450, numbers of the form (4^k-1)/3, for k = 2, 4, 5, 7, 8, 10, ... (note that k a multiple of 3 does not appear).
A124477 \ {0,1} is a subset: for these n, 3n+1 = 2^(p-3) with p > 3 prime, whence also n !== 0 (mod 3). - M. F. Hasler, Oct 16 2018
These are exactly the odd non-multiples of 3 such 3n+1 = 2^m for some m, i.e., n = (2^m-1)/3. This is possible iff m = 2k, so we get n = (4^k-1)/3. Then n == 0 (mod 3) <=> 4^k == 1 (mod 9) <=> k == 0 (mod 3) <=> k not in A001651. This yields the FORMULA. (Multiples of 3 are excluded because the original definition implied that the terms are in the Collatz-orbit of another odd number, i.e., of the form n = (3x+1)/2^r, which is impossible for x a multiple of 3.) - M. F. Hasler, Oct 16 2018
From Wolfdieter Lang, Jan 14 2022: (Start)
a(n) mod 8 = 5. As subsequence of A002450 for n >= 1.
{a(n) mod 6} == repeat{5, 1}. See the first comment, and the periodicity modulo 6 of A002450 for n >= 1.
{a(n) mod 72} == repeat{5, 13, 53, 61, 29, 37}. Proof by induction: First with the bisection formulas, a(1+2*k) = (4^(2+3*k) - 1)/3 and a(2+2*k) = (4^(3*k+4) - 1)/3, for k >= 0, then trisection, using (4^9 - 1)/3 = 873819 = 9*9709. (End)

T. D. Noe, Table of n, a(n) for n = 1..100
Index entries for linear recurrences with constant coefficients, signature (1,64,-64).

Cf. A001651, A002450, A124477, A198584.

[4^(3*n  div 2 + 1) div 3: n in [1..25]]; // Vincenzo Librandi, Oct 20 2018

e = 19; ex = Complement[Range[2,3*e], 3*Range[e]]; (4^ex - 1)/3
(* Second program: *)
Rest@ Map[(4^# - 1)/3 &, LinearRecurrence[{1, 1, -1}, {1, 2, 4}, 21]] (* Michael De Vlieger, Oct 17 2018 *)

is(n)=gcd(n,6)==1&&(n=3*n+1)>>valuation(n,2)==1 \\ M. F. Hasler, Oct 16 2018

A198586(n)=4^(3*n\2+1)\3 \\ M. F. Hasler, Oct 16 2018

Vec(x*(5 + 80*x - 64*x^2) / ((1 - x)*(1 - 8*x)*(1 + 8*x)) + O(x^20)) \\ Colin Barker, Jan 17 2020

a(n) = (4^A001651(n+1) - 1)/3. - M. F. Hasler, Oct 16 2018
From Colin Barker, Jan 17 2020: (Start)
G.f.: x*(5 + 80*x - 64*x^2) / ((1 - x)*(1 - 8*x)*(1 + 8*x)).
a(n) = a(n-1) + 64*a(n-2) - 64*a(n-3) for n>3.
a(n) = (-1 + (-8)^n + 3*8^n) / 3.
(End)

Definition corrected by M. F. Hasler, Oct 16 2018

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

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A198586 a(n) = (4^A001651(n+1) - 1)/3: numbers (4^k-1)/3 for k > 1, not multiples of 3.

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