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%I #23 Mar 11 2022 12:39:50
%S 5,17,17,53,29,53,41,161,53,89,65,161,77,125,89,485,101,161,113,269,
%T 125,197,137,485,149,233,161,377,173,269,185,1457,197,305,209,485,221,
%U 341,233,809,245,377,257,593,269,413,281,1457,293,449,305,701,317,485
%N a(n) is the first maximum reached by iterating the reduced Collatz function R on 4n-1: a(n) = R^s(4n-1), where R(k) = A139391(k) and s the number of iterations required.
%C Iterating R on 4n-1 (n>=1) starts with an increasing trajectory before reaching the first maximum. However, iterating R on 4n-3 (n>=1) starts with a decreasing trajectory before reaching 1 or the first minimum.
%F a(n) = 4*n*(3/2)^s - 1, where s = A001511(n).
%F a(n) == 5 (mod 12).
%F For s >= 1 and m >= 0, a(2^s*m+2^(s-1)) = 2*(3^s)*(2m+1) - 1. For example, a(2m+1) = 12m+5 = A017581(m); a(4m+2) = 36m+17; and a(8m+4) = 108m+53.
%e For n = 1, iterating R on 4n-1=3 gives 3->5->1, in which the first maximum is 5, and thus a(0) = 5.
%e For n = 8, iterating R on 4n-1=31 gives 31->47->71->107->161->121->91->137->103->155->233->175...->23->35->53->5->1, in which the first maximum is 161, and thus a(8) = 161.
%o (Python)
%o def A351974(n): s = (n&-n).bit_length(); return 4*n*3**s//2**s - 1
%o (PARI) a(n) = my(s=valuation(n,2)); n>>(s-1)*3^(s+1) - 1; \\ _Kevin Ryde_, Feb 28 2022
%Y Cf. A004767, A017581, A001511, A139391, A075684.
%K nonn,easy
%O 1,1
%A _Ya-Ping Lu_, Feb 26 2022
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