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A198584
Odd numbers producing 3 odd numbers in the Collatz (3x+1) iteration.
17
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
1,1
COMMENTS
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.
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
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
EXAMPLE
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.
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
PROG
(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+(c<s)+(c>4*s+1)+(c>5*s+1))+5
e = isqrt(d)
f = e-1-( (d-e*e) >> 1 )
r = ((((8<<e)-(1<<f))//3)-1)//3
print(r, end =", ") # 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<<a)-(1<<b))//3-1)//3 , end=", ") # André Hallqvist, Aug 14 2019
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).
Sequence in context: A346409 A082376 A065059 * A346382 A342815 A072197
KEYWORD
nonn,easy
AUTHOR
T. D. Noe, Oct 28 2011
STATUS
approved