A353215 a(n) is the result of n applications of the function f on n, where f(x) = floor((3*x - 1)/2) (A001651).

0, 1, 2, 7, 14, 29, 50, 95, 164, 304, 475, 824, 1370, 2297, 3598, 5906, 9370, 15586, 23848, 39227, 61448, 98114, 151318, 240098, 378290, 599105, 916738, 1454537, 2261948, 3543307, 5448094, 8486453, 13088486, 20669311, 31151588, 49081505, 75209263, 116597314

Offset

0,3

Links

Table of n, a(n) for n=0..37.

Formula

a(n) = f^n(n) where f(n) = floor((3*n - 1)/2) = A001651(n).

Example

a(0) = f^0 (0) = 0 (f not applied at all);
a(1) = f^1 (1) = f(1) = floor((3*1 - 1)/2) = 1;
a(2) = f^2 (2) = f(f(2)) = floor((3*f(2) - 1)/2) = floor((3*floor((3*2 - 1)/2) - 1)/2) = 2.

Maple

a:= n-> (f-> (f@@n)(n))(t-> floor((3*t-1)/2)):
seq(a(n), n=0..20);

Prog

(C++)
#include <iostream>
using namespace std;
// Think of unsigned int as a natural number
unsigned int f(unsigned int n) {
    return (3*n - 1)/2;
}
unsigned int a(unsigned int pow, unsigned int n) {
    if (pow == 0) return n;
    else return a(pow-1, f(n));
}
int main() {
    for (unsigned int n(0); n <= 20; ++n)
        cout << a(n, n) << " ";
    return 0;
}
(Python)
def f(n):
    return (3*n - 1)//2;
def a(pow, n):
    if (pow == 0): return n;
    else: return a(pow-1, f(n));
l = [a(n, n) for n in range(21)];
(Python)
from functools import reduce
def A353215(n): return reduce(lambda x, _ : (3*x-1)//2, range(n), n) # Chai Wah Wu, May 07 2022

Crossrefs

Cf. A001651 (step), A353220.

Sequence in context: A005998 A122751 A152944 * A018437 A286857 A289268

Adjacent sequences: A353212 A353213 A353214 * A353216 A353217 A353218

Keyword

nonn

Author

Yves Daaboul, May 01 2022

Status

approved