login
A351974
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.
3
5, 17, 17, 53, 29, 53, 41, 161, 53, 89, 65, 161, 77, 125, 89, 485, 101, 161, 113, 269, 125, 197, 137, 485, 149, 233, 161, 377, 173, 269, 185, 1457, 197, 305, 209, 485, 221, 341, 233, 809, 245, 377, 257, 593, 269, 413, 281, 1457, 293, 449, 305, 701, 317, 485
OFFSET
1,1
COMMENTS
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.
FORMULA
a(n) = 4*n*(3/2)^s - 1, where s = A001511(n).
a(n) == 5 (mod 12).
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.
EXAMPLE
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.
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.
PROG
(Python)
def A351974(n): s = (n&-n).bit_length(); return 4*n*3**s//2**s - 1
(PARI) a(n) = my(s=valuation(n, 2)); n>>(s-1)*3^(s+1) - 1; \\ Kevin Ryde, Feb 28 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ya-Ping Lu, Feb 26 2022
STATUS
approved