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A078421 Least positive integer solution k to h(k) = n h(k-1), where h(n) is the length of n, f(n), f(f(n)), ...., 1 in the Collatz (or 3x + 1) problem. (The earliest "1" is meant.) 0
13, 2, 105, 3, 9, 1742, 1961, 1705, 161, 11945, 68718937, 36453025, 411755350, 3222597649, 2808250369, 14913078 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

1. Recall that f(n) = n/2 if n is even; = 3n + 1 if n is odd. 2. Problem: Is a(n) defined for all n, that is, does a positive integer solution k to h(k) = n h(k-1) always exist? 3. If a(11) is defined, then a(11) > 3 x 10^5.

a(11)>2^22 - Michel ten Voorde Jun 20 2003

LINKS

Table of n, a(n) for n=1..16.

EXAMPLE

n, f(n), f(f(n)), ...., 1 for n = 105, 104, respectively, are: 105, 316, 158, 79, 238, 119, 358, 179, 538, 269, 808, 404, 202, 101, 304, 152, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1; 104, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, of lengths 39 = 3 x 13 and 13, respectively. Hence k = 105 is a solution to h(k) = 3 h(k-1) and it is also the smallest solution. This implies a(3) = 105.

MATHEMATICA

h[n_] := Module[{a, i}, i = n; a = 1; While[i > 1, a = a + 1; i = f[i]]; a]; a = {}; For[k = 1, k <= 30, k++, j = 2; While[ h[j] != k h[j - 1], j = j + 1]; a = Append[a, j]]; a

CROSSREFS

Sequence in context: A176593 A031066 A154355 * A185808 A178548 A098222

Adjacent sequences:  A078418 A078419 A078420 * A078422 A078423 A078424

KEYWORD

more,nonn

AUTHOR

Joseph L. Pe, Dec 29 2002

EXTENSIONS

a(11)-a(16) from Donovan Johnson, Nov 14 2010

STATUS

approved

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Last modified January 29 01:42 EST 2020. Contains 331328 sequences. (Running on oeis4.)