Problem 51715. Iterate the sum of divisors and totient

Cody Problem 46898 deals with the sum of divisors function, denoted by sigma(n), while Cody Problem 656 deals with the totient function, denoted by phi(n). The sum of divisors is straightforward: for example, sigma(12) = 1+2+3+4+6+12 = 28. The totient of n counts the numbers less than n that are relatively prime to n. For example phi(12) = 4 because the greatest common divisor of 12 and four numbers (1, 5, 7, 11) is 1.
What happens if you repeatedly apply the two functions, starting with the sum of divisors and alternating? For example, start with 7. Then
sigma(7) = 8, phi(8) = 4, sigma(4) = 7, phi(7) = 6, sigma(6) = 12, phi(12) = 4, etc.
and the pattern 7, 6, 12, 4 will repeat.
Oscillating behavior is plausible because the sum of divisors is always greater than n, and the totient is always smaller than n. Furthermore, because the totient has a minimum value and the sum of divisors has a maximum value, with enough iterations the sequence would have to hit a repeating pattern.
Write a function that takes an initial seed and returns the repeating pattern and the index of the sequence where the pattern begins. With an initial seed of 7, the sequence would be 7, 8, 4, 7, 6, 12, 4, 7, 6, 12,… Therefore, the repeating pattern is [7 6 12 4] and the start index is 3.

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62.5% Correct | 37.5% Incorrect
Last Solution submitted on Apr 24, 2024

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