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

0,2

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

Links

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

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)

Example

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.

Prog

(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
(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

Crossrefs

Cf. A010872, A349876, A353311, A353313.

Sequence in context: A219221 A249510 A324219 * A196815 A260120 A221234

Adjacent sequences: A349874 A349875 A349876 * A349878 A349879 A349880

Keyword

nonn,easy

Author

Nicholas Drozd, Dec 03 2021

Extensions

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

Status

approved