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A198584
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Odd numbers producing 3 odd numbers in the Collatz (3x+1) iteration.
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17
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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MATHEMATICA
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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
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PROG
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(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
(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
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CROSSREFS
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Cf. A062053 (numbers producing 3 odds in their Collatz iteration).
Cf. A092893 (least number producing n odd numbers).
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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