%I #21 Aug 31 2021 16:47:54
%S 113,453,1813,7253,7281,29013,29125,116053,116501,464213,466005,
%T 466033,1856853,1864021,1864133,7427413,7456085,7456533,29709653,
%U 29824341,29826133,29826161,118838613,119297365,119304533,119304645,475354453,477189461,477218133
%N Numbers of the form (2^(2*j + 6*k + 10) - 2^(2*j + 2) - 3)/9, with j,k >= 0.
%C Sequence is a subsequence of A198584. When any term is iterated using the Collatz function, the last odd integer in the trajectory before 1 is of the form (4^(3*m + 4) - 1)/3.
%H Satya Das, <a href="https://www.researchgate.net/publication/354253993_3X1_PROBLEM_A_CONTINUOUS_EXTENSION_OF_THE_SPEEDED_UP_COLLATZ_MAP">Extension of speeded up Collatz map to the real line</a>
%t Take[Sort[Flatten[Table[(2^(2n1+6n2+10) - 2^(2n1+2) - 3)/9, {n1, 0, 20}, {n2, 0, 20}]]], 50]
%o (Python)
%o seq=[]
%o for n1 in range(20):
%o for n2 in range(20):
%o n=(2**(2*n1+6*n2+10) - 2**(2*n1+2) - 3)/9
%o seq.append(n)
%o seq.sort()
%o print(seq[0:50])
%Y Union with A342815 gives A198584.
%K nonn,easy
%O 1,1
%A _Satya Das_, Mar 22 2021