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A228871
Odd numbers producing 3 out-of-order odd numbers in the Collatz (3x+1) iteration.
5
3, 227, 14563, 932067, 59652323, 3817748707, 244335917283, 15637498706147, 1000799917193443, 64051194700380387, 4099276460824344803, 262353693492758067427, 16790636383536516315363, 1074600728546337044183267, 68774446626965570827729123
OFFSET
1,1
COMMENTS
Sequence A198584 gives the first term of the Collatz sequence having exactly 3 odd numbers. This sequence is the subset of A198584 for which the second odd number is larger than the first. The second odd number is (2^(6*n - 2) - 1)/3, which always occurs as the third term of the sequence.
{a(n) mod 6} = {repeat(3, 5, 1)}, and a(n) mod 8 = 3 for all n. Proof from the formula of a(n) in terms of A198586 given below, using the modulo 72 congruence of the odd indexed part of A198586 given there. - Wolfdieter Lang, Jan 14 2022
FORMULA
a(n) = (64^n/2 - 5)/9. - Alois P. Heinz, Dec 08 2021
From Wolfdieter Lang, Jan 12 2022: (Start)
a(n) = (2*A198586(2*n-1) - 1)/3. See the Mathematica program.
G.f.: x*(3 + 32*x)/((1 - x)*(1 - 64*x)). (End)
EXAMPLE
The number 3 has the Collatz iteration {3, 10, 5, 16, 8, 4, 2, 1}, which has three out-of-order odd numbers {3, 5, 1}.
MATHEMATICA
Table[(2*(2^(6*n - 2) - 1)/3 - 1)/3, {n, 15}]
PROG
(PARI) a(n)=4^(3*n-1)\3*2\3 \\ Charles R Greathouse IV, Mar 11 2017
CROSSREFS
Cf. A198584 (Collatz iterations having 3 odd numbers).
Cf. A228872 (Collatz iterations producing 3 in-order odd numbers).
Cf. A198586.
Sequence in context: A241098 A254157 A131493 * A195500 A099426 A332123
KEYWORD
nonn,easy
AUTHOR
T. D. Noe, Sep 12 2013
STATUS
approved