A323832 Start with n and repeatedly double it and apply the "repeatedly delete any run of identical digits" operation described in A323830; a(n) is the number of steps needed to reach one of 0, 1, or 5, or -1 if none of these three numbers is ever reached.

0, 0, 19, 12, 18, 0, 11, 23, 17, 4, 19, 1, 10, 29, 22, 32, 16, 5, 3, 47, 18, 15, 1, 20, 9, 2, 28, 26, 21, 13, 31, 24, 15, 1, 4, 23, 2, 18, 46, 21, 17, 51, 14, 15, 1, 24, 19, 2, 8, 10, 1, 33, 27, 24, 25, 1, 20, 19, 12, 18, 30, 1, 23, 7, 14, 29, 5, 20, 3

Offset: 0

Author

N. J. A. Sloane, Feb 03 2019

Keywords

• nonn
• base

Comments

Conjecture: every number will eventually reach one of 0, 1, or 5.

From Chai Wah Wu, Feb 04 2019, Feb 13 2019: (Start)

Conjecture is true for n < 10^10.

1604466 takes 115 steps to reach 5 and is the largest value for a(n) for n < 10^7.

91070713 takes 121 steps to reach 5 and is the largest value for a(n) for n < 10^8.

126591463 and 801282051 both take 128 steps to reach 5 and this is the largest value for a(n) for n < 10^9.

The numbers 1582393271, 1582393293, 4645106705 all take 131 steps to reach 5 and this is the largest value for a(n) for n < 10^10.

(End)

Examples

			Starting with 2, the trajectory is 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 636, 1272, 25, 50, 1, reaching 1 in 20 steps, so a(2) = 20.
3 reaches 1 in 12 steps: 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 61, 1, so a(3) = 12.
10 reaches 5 in 19 steps: 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 6360, 12720, 250, 5, so a(10) = 19.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A321801, A321802, A323830, A323830.

Programs

Python

from re import split
def A321801(n):
    return int('0'+''.join(d for d in split('(0+)|(1+)|(2+)|(3+)|(4+)|(5+)|(6+)|(7+)|(8+)|(9+)',str(n)) if d != '' and d != None and len(d) == 1))
def f(n):
    x = 2*n
    y = A321801(x)
    while x != y:
        x, y = y, A321801(y)
    return x
def A323832(n):
    mset, m, c = set(), n, 0
    while True:
        if m == 1 or m == 0 or m == 5:
            return c
        m = f(m)
        if m in mset:
            return -1
        mset.add(m)
        c += 1 # Chai Wah Wu, Feb 04 2019, Feb 11 2019