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.

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.

Links

Table of n, a(n) for n=1..54.

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

Cf. A004767, A017581, A001511, A139391, A075684.

Sequence in context: A231710 A316548 A275629 * A364935 A249589 A040153

Adjacent sequences: A351971 A351972 A351973 * A351975 A351976 A351977

Keyword

nonn,easy

Author

Ya-Ping Lu, Feb 26 2022

Status

approved