cp's OEIS Frontend

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.

%I A323832 #38 Feb 13 2019 17:24:17
%S A323832 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,
%T A323832 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,
%U A323832 33,27,24,25,1,20,19,12,18,30,1,23,7,14,29,5,20,3
%N 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.
%C A323832 Conjecture: every number will eventually reach one of 0, 1, or 5.
%C A323832 From _Chai Wah Wu_, Feb 04 2019, Feb 13 2019: (Start)
%C A323832 Conjecture is true for n < 10^10.
%C A323832 1604466 takes 115 steps to reach 5 and is the largest value for a(n) for n < 10^7.
%C A323832 91070713 takes 121 steps to reach 5 and is the largest value for a(n) for n < 10^8.
%C A323832 126591463 and 801282051 both take 128 steps to reach 5 and this is the largest value for a(n) for n < 10^9.
%C A323832 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.
%C A323832 (End)
%H A323832 Chai Wah Wu, <a href="/A323832/b323832.txt">Table of n, a(n) for n = 0..10000</a>
%e A323832 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.
%e A323832 3 reaches 1 in 12 steps: 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 61, 1, so a(3) = 12.
%e A323832 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.
%o A323832 (Python)
%o A323832 from re import split
%o A323832 def A321801(n):
%o A323832     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))
%o A323832 def f(n):
%o A323832     x = 2*n
%o A323832     y = A321801(x)
%o A323832     while x != y:
%o A323832         x, y = y, A321801(y)
%o A323832     return x
%o A323832 def A323832(n):
%o A323832     mset, m, c = set(), n, 0
%o A323832     while True:
%o A323832         if m == 1 or m == 0 or m == 5:
%o A323832             return c
%o A323832         m = f(m)
%o A323832         if m in mset:
%o A323832             return -1
%o A323832         mset.add(m)
%o A323832         c += 1 # _Chai Wah Wu_, Feb 04 2019, Feb 11 2019
%Y A323832 Cf. A321801, A321802, A323830, A323830.
%K A323832 nonn,base
%O A323832 0,3
%A A323832 _N. J. A. Sloane_, Feb 03 2019