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%I #29 Jan 20 2022 15:55:57
%S 3,227,14563,932067,59652323,3817748707,244335917283,15637498706147,
%T 1000799917193443,64051194700380387,4099276460824344803,
%U 262353693492758067427,16790636383536516315363,1074600728546337044183267,68774446626965570827729123
%N Odd numbers producing 3 out-of-order odd numbers in the Collatz (3x+1) iteration.
%C 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.
%C {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
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (65,-64).
%F a(n) = (64^n/2 - 5)/9. - _Alois P. Heinz_, Dec 08 2021
%F From _Wolfdieter Lang_, Jan 12 2022: (Start)
%F a(n) = (2*A198586(2*n-1) - 1)/3. See the Mathematica program.
%F G.f.: x*(3 + 32*x)/((1 - x)*(1 - 64*x)). (End)
%e 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}.
%t Table[(2*(2^(6*n - 2) - 1)/3 - 1)/3, {n, 15}]
%o (PARI) a(n)=4^(3*n-1)\3*2\3 \\ _Charles R Greathouse IV_, Mar 11 2017
%Y Cf. A198584 (Collatz iterations having 3 odd numbers).
%Y Cf. A228872 (Collatz iterations producing 3 in-order odd numbers).
%Y Cf. A198586.
%K nonn,easy
%O 1,1
%A _T. D. Noe_, Sep 12 2013
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