%I #36 Oct 11 2022 14:11:39

%S 3,13,53,113,213,227,453,853,909,1813,3413,3637,7253,7281,13653,14549,

%T 14563,29013,29125,54613,58197,58253,116053,116501,218453,232789,

%U 233013,464213,466005,466033,873813,931157,932053,932067,1856853,1864021,1864133

%N Odd numbers producing 3 odd numbers in the Collatz (3x+1) iteration.

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

%C 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

%H T. D. Noe, <a href="/A198584/b198584.txt">Table of n, a(n) for n = 1..1009</a>

%F 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

%e 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 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

%o (Python)

%o # get n-th term in sequence

%o def isqrt(n):

%o i=0

%o while(i*i<=n):

%o i+=1

%o return i-1

%o for n in range (200):

%o s = isqrt(3*n)//3

%o a = s*3

%o b = (a*a)//3

%o c = n-b

%o d = 4*(n*3+a+(c<s)+(c>4*s+1)+(c>5*s+1))+5

%o e = isqrt(d)

%o f = e-1-( (d-e*e) >> 1 )

%o r = ((((8<<e)-(1<<f))//3)-1)//3

%o print(r,end =", ") # _André Hallqvist_, Jul 25 2019

%o (Python)

%o # just prints the sequence

%o for a in range (5,100,1):

%o for b in range(a-8+4*(a&1),0,-6):

%o print(( ((1<<a)-(1<<b))//3-1)//3 ,end=",") # _André Hallqvist_, Aug 14 2019

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

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

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

%Y Cf. A228871, A228872.

%K nonn,easy

%O 1,1

%A _T. D. Noe_, Oct 28 2011