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%I #193 Jan 12 2023 02:09:53
%S 3,7,11,15,23,27,31,39,47,59,63,71,79,91,95,103,111,123,127,155,159,
%T 167,175,191,199,207,219,223,231,239,251,255,283,287,303,319,327,347,
%U 359,367,383,411,415,423,447,463,479,487,495,507,511,539,543,559,575
%N Starting values of the reduced Collatz function (A014682) where 2 to the power of the "dropping time" is greater than the starting value.
%C a(n) is the lowest positive starting value of the reduced Collatz function such that all starting values (>1) that are congruent to a(n) (mod 2^d) have the same dropping time (d). The dropping time here counts the (3x+1)/2 and the x/2 steps as listed in A126241. A number is included in this sequence if 2^A126241(a(n)) > a(n).
%C Starting values that produce new record dropping times as listed in A060412 are necessarily a subset of this sequence.
%C If at least one iteration is carried out before checking that the absolute iterated value has become less than or equal to the absolute starting value, then a(n) is the lowest positive starting value such that all starting values (positive, zero or negative) that are congruent to a(n) (mod 2^d) have the same dropping time (d). Defined like this, the sequence would start with 0, 1, 3, 7.
%C For k>0, A076227(k) is the number of terms between 2^k and 2^(k+1)-1. - _Ruud H.G. van Tol_, Dec 18 2022
%C All terms are congruent to 3 (mod 4) since any 1 (mod 4) has dropping time A126241(4k+1) = 2, for k>=1. - _Ruud H.G. van Tol_, Jan 11 2023
%e 3 is in this sequence because the dropping time starting with 3 is A126241(3) = 4 and 2^4 > 3.
%o (PARI) is(t)= if(t<3||3!=t%4,0,my(x=t, d=0); until(x<=t, if(x%2, x=(x*3+1)/2, x/=2); d++); 2^d>t); \\ updated by _Ruud H.G. van Tol_, Jan 10 2023
%Y Cf. A014682, A060412, A076227, A126241, A177789, A239126.
%K nonn
%O 1,1
%A _K. Spage_, Aug 20 2014
%E Offset 1 from _Ruud H.G. van Tol_, Jan 10 2023
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