A349877 a(n) is the number of times the map x -> A353314(x) needs to be applied to n to reach a multiple of 3, or -1 if the trajectory never reaches a multiple of 3.

0, 2, 14, 0, 1, 13, 0, 4, 1, 0, 12, 3, 0, 1, 3, 0, 4, 1, 0, 11, 2, 0, 1, 2, 0, 2, 1, 0, 2, 3, 0, 1, 3, 0, 10, 1, 0, 4, 5, 0, 1, 7, 0, 3, 1, 0, 3, 2, 0, 1, 2, 0, 2, 1, 0, 2, 4, 0, 1, 9, 0, 3, 1, 0, 3, 4, 0, 1, 5, 0, 6, 1, 0, 4, 2, 0, 1, 2, 0, 2, 1, 0, 2, 7, 0, 1, 4, 0, 6, 1, 0, 6, 3, 0, 1, 3, 0, 5, 1, 0, 8, 2, 0

Offset: 0

Author

Nicholas Drozd, Dec 03 2021

Keywords

• nonn
• easy

Comments

Equally, number of iterations of A353313 needed to reach a multiple of 3, or -1 if no multiple of 3 is ever reached. - Antti Karttunen, Apr 14 2022

Examples

			a(1) = 2 : 1 -> 4 -> 9 (as it takes two applications of A353314 to reach a multiple of three),
a(2) = 14 : 2 -> 5 -> 10 -> 19 -> 34 -> 59 -> 100 -> 169 -> 284 -> 475 -> 794 -> 1325 -> 2210 -> 3685 -> 6144
a(3) = 0 : 3 (as the starting point 3 is already a multiple of 3).
a(4) = 1 : 4 -> 9
a(7) = 4 : 7 -> 14 -> 25 -> 44 -> 75.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..19683

Crossrefs

Cf. A010872, A349876, A353311, A353313.

Programs

PARI

A353314(n) = { my(r=(n%3)); if(!r,n,((5*((n-r)/3)) + r + 3)); };
A349877(n) = { my(k=0); while(n%3, k++; n = A353314(n)); (k); }; \\ Antti Karttunen, Apr 14 2022

Python

import itertools
def f(n):
    for i in itertools.count():
        quot, rem = divmod(n, 3)
        if rem == 0:
            return i
        n = (5 * quot) + rem + 3

Formula

From Antti Karttunen, Apr 14 2022: (Start)

If A010872(n) = 0 then a(n) = 0, otherwise a(n) = 1 + a(A353314(n)).

a(n) < A353311(n) for all n.

(End)

Extensions

Definition corrected and more terms from Antti Karttunen, Apr 14 2022