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

Author

Ya-Ping Lu, Feb 26 2022

Keywords

• nonn
• easy

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.

Examples

			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.
		

Crossrefs

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

Programs

PARI

a(n) = my(s=valuation(n,2)); n>>(s-1)*3^(s+1) - 1; \\ Kevin Ryde, Feb 28 2022

Python

def A351974(n): s = (n&-n).bit_length(); return 4*n*3**s//2**s - 1

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.