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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. 6
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

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)

LINKS

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

EXAMPLE

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.

PROG

(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

CROSSREFS

Cf. A321801, A321802, A323830, A323830.

Sequence in context: A083156 A040344 A261717 * A306384 A196188 A089294

Adjacent sequences:  A323829 A323830 A323831 * A323833 A323834 A323835

KEYWORD

nonn,base

AUTHOR

N. J. A. Sloane, Feb 03 2019

STATUS

approved

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Last modified September 20 10:32 EDT 2019. Contains 327229 sequences. (Running on oeis4.)