The Collatz conjecture is a conjecture in mathematics named after Lothar Collatz, who first proposed it in 1937. The conjecture is also known as the 3n + 1 conjecture, the Ulam conjecture (after Stanislaw Ulam), Kakutani's problem (after Shizuo Kakutani), the Thwaites conjecture (after Sir Bryan Thwaites), Hasse's algorithm (after Helmut Hasse), or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers, or as wondrous numbers.
Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process (which has been called "Half Or Triple Plus One", or HOTPO[4]) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has also been called oneness.
Paul Erdős said, allegedly, about the Collatz conjecture: "Mathematics is not yet ripe for such problems" and also offered $500 for its solution.
Statement of the problem:
Consider the following operation on an arbitrary positive integer:
If the number is even, divide it by two.
If the number is odd, triple it and add one.
In modular arithmetic notation, define the function f as follows:
Numbers from 1 to 9999 and their corresponding total stopping time.
Now, form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next.
In notation:
(that is: is the value of applied to recursively times; ).
The Collatz conjecture is: This process will eventually reach the number 1, regardless of which positive integer is chosen initially.
That smallest i such that ai = 1 is called the total stopping time of n.[9] The conjecture asserts that every n has a well-defined total stopping time. If, for some n, such an i doesn't exist, we say that n has infinite total stopping time and the conjecture is false.
If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence which does not contain 1. Such a sequence might enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found.
–
Sayan Chaudhuri
·
4 years, 3 months ago

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Examples::::::::
For instance, starting with n = 6, one gets the sequence 6, 3, 10, 5, 16, 8, 4, 2, 1.
n = 11, for example, takes longer to reach 1: 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
The sequence for n = 27, listed and graphed below, takes 111 steps, climbing to 9232 before descending to 1.
{ 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 }
Numbers with a total stopping time longer than any smaller starting value form a sequence beginning with:
1, 2, 3, 6, 7, 9, 18, 25, 27, 54, 73, 97, … (sequence A006877 in OEIS).
The longest progression for any initial starting number less than 100 million is 63,728,127, which has 949 steps. For starting numbers less than 1 billion it is 670,617,279, with 986 steps, and for numbers less than 10 billion it is 9,780,657,630, with 1132 steps....
The powers of two converge to one quickly because is halved times to reach one, and is never increased.
–
Sayan Chaudhuri
·
4 years, 3 months ago

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TopNewestThe Collatz conjecture is a conjecture in mathematics named after Lothar Collatz, who first proposed it in 1937. The conjecture is also known as the 3n + 1 conjecture, the Ulam conjecture (after Stanislaw Ulam), Kakutani's problem (after Shizuo Kakutani), the Thwaites conjecture (after Sir Bryan Thwaites), Hasse's algorithm (after Helmut Hasse), or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers, or as wondrous numbers. Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process (which has been called "Half Or Triple Plus One", or HOTPO[4]) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has also been called oneness. Paul Erdős said, allegedly, about the Collatz conjecture: "Mathematics is not yet ripe for such problems" and also offered $500 for its solution. Statement of the problem: Consider the following operation on an arbitrary positive integer: If the number is even, divide it by two. If the number is odd, triple it and add one. In modular arithmetic notation, define the function f as follows: Numbers from 1 to 9999 and their corresponding total stopping time. Now, form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next. In notation: (that is: is the value of applied to recursively times; ). The Collatz conjecture is: This process will eventually reach the number 1, regardless of which positive integer is chosen initially. That smallest i such that ai = 1 is called the total stopping time of n.[9] The conjecture asserts that every n has a well-defined total stopping time. If, for some n, such an i doesn't exist, we say that n has infinite total stopping time and the conjecture is false. If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence which does not contain 1. Such a sequence might enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found. – Sayan Chaudhuri · 4 years, 3 months ago

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Examples:::::::: For instance, starting with n = 6, one gets the sequence 6, 3, 10, 5, 16, 8, 4, 2, 1. n = 11, for example, takes longer to reach 1: 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. The sequence for n = 27, listed and graphed below, takes 111 steps, climbing to 9232 before descending to 1. { 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 } Numbers with a total stopping time longer than any smaller starting value form a sequence beginning with: 1, 2, 3, 6, 7, 9, 18, 25, 27, 54, 73, 97, … (sequence A006877 in OEIS). The longest progression for any initial starting number less than 100 million is 63,728,127, which has 949 steps. For starting numbers less than 1 billion it is 670,617,279, with 986 steps, and for numbers less than 10 billion it is 9,780,657,630, with 1132 steps.... The powers of two converge to one quickly because is halved times to reach one, and is never increased. – Sayan Chaudhuri · 4 years, 3 months ago

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