A340801 -id:A340801 - cp's OEIS frontend

A385613 Number of steps that n requires to reach 0 under the map: x-> x^2 - 1 if x is an odd prime, floor(x/3) if x is even, otherwise x - 1. a(n) = -1 if 0 is never reached.

0, 1, 1, 3, 2, 4, 2, 7, 2, 3, 4, 9, 3, 6, 3, 4, 5, 8, 3, 10, 3, 4, 8, 14, 3, 4, 3, 4, 4, 10, 5, 15, 5, 6, 10, 11, 4, 10, 4, 5, 7, 8, 4, 14, 4, 5, 5, 10, 6, 7, 6, 7, 9, 15, 4, 5, 4, 5, 11, 9, 4, 18, 4, 5, 5, 6, 9, 10, 9, 10, 15, 9, 4, 22, 4, 5, 5, 6, 4, 11, 4

Offset: 0

Ya-Ping Lu, Jul 04 2025

sign

n = 122827 is the smallest starting number that ends up in a loop. The loop contains 33 elements: 122827 -> 15086471928 -> 5028823976 -> 1676274658 -> 558758219 -> 558758218 -> 186252739 -> 186252738 -> 62084246 -> 20694748 -> 6898249 -> 47585839266000 -> 15861946422000 -> 5287315474000 -> 1762438491333 -> 1762438491332 -> 587479497110 -> 195826499036 -> 65275499678 -> 21758499892 -> 7252833297 -> 7252833296 -> 2417611098 -> 805870366 -> 268623455 -> 268623454 -> 89541151 -> 89541150 -> 29847050 -> 9949016 -> 3316338 -> 1105446 -> 368482 -> 122827.
No other loop is found for n < 164901049.
Conjecture: a(n) = -1 occurs only when starting number n runs into a loop.

			a(5) = 4 because iterating the map on n = 5 results in 0 in 4 steps: 5 -> 5^2-1=24 -> floor(24/3)=8 -> floor(8/3)=2 -> floor(2/3)=0.
a(9949031) = -1 because iterating the map on n = 9949031 ends up in the 33-member loop in 5 steps: 9949031 -> 9949030 -> 3316343 -> 3316342 -> 1105447 -> 1105446.

Cf. A339991, A340801, A384713.

from sympy import isprime
def A385613(n):
    S = {n}
    while n != 0:
        n = n//3 if n%2 == 0 else n*n - 1 if isprime(n) else n - 1
        if n in S: return -1
        S.add(n)
    return len(S) - 1

A341742 Nodes read by depth in a binary tree defined as: Root = 1; an even node N has a left child N + 1 if N + 1 is not a prime, and an odd node N has a left child sqrt(N + 2) if sqrt(N + 2) is a prime; the right child of a node N is 2*N.

Original entry on oeis.org

1, 2, 4, 8, 9, 16, 18, 32, 36, 33, 64, 72, 66, 65, 128, 144, 132, 130, 129, 256, 145, 288, 133, 264, 260, 258, 512, 290, 289, 576, 266, 265, 528, 261, 520, 259, 516, 513, 1024, 291, 580, 578, 1152, 267, 532, 530, 529, 1056, 522, 1040, 518, 517, 1032, 1026

Offset: 1

Ya-Ping Lu, Feb 18 2021

nonn

Let d be the depth of a node N in the binary tree and f be the map of A340801. The d-th iteration of map A340801 on N gives 1, or f^d(N) = 1.
If Conjectures 1 and 2 made in A340801 hold, the sequence contains all positive integers and each integer appears once in the sequence.
The first odd prime does not appear until d reaches 30 and the first five odd primes appearing in the sequence are:
        n     a(n)     d
  -------    -----    --
   140735     4099    30
   151872     1543    31
  1574120     8689    36
  1841645     2917    36
  2111465    32771    36
The first two odd primes less than 100 appear in the binary tree are 17 at d = 4426 and 71 at d = 4421.

			The binary tree for depths up to 9 is given below.
  1
   \
    2
     \
      4
       \
        8
     /    \
   9       16
    \        \
    18        32
     \       /  \
     36    33    64
      \     \    / \
      72    66  65  128
       \     \   \   / \
      144   132 130 129 256
      / \   / \   \   \   \
   145 288 133 264 260 258 512

Cf. A340801, A006577, A340008, A339991, A340419.

from sympy import isprime
from math import sqrt
def children(N):
    C = []
    if N%2 == 0:
        if isprime(N + 1) == 0: C.append(N+1)
    else:
        p1 = sqrt(N + 2.0); p2 = int(p1 + 0.5)
        if p2**2 == N + 2 and isprime(p2) == 1: C.append(p2)
    C.append(2*N)
    return C
L_last = [1]; print(L_last)
for d in range(1, 18):
    L_1 = []
    for i in range(0, len(L_last)):
        C_i = children(L_last[i])
        for j in range(0, len(C_i)): L_1.append(C_i[j])
    print(L_1); L_last = L_1

cp's OEIS Frontend

A385613 Number of steps that n requires to reach 0 under the map: x-> x^2 - 1 if x is an odd prime, floor(x/3) if x is even, otherwise x - 1. a(n) = -1 if 0 is never reached.

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