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Narcissistic/Plus Perfect/Armstrong 
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#1
Feb213, 03:18 PM

P: 531

I found reference to a function described as taking the digits of a natural number and summing their third powers, iterating the process. The claim was that 1/3 of all natural numbers will terminate the process in 153.
I did this initiating the process with the first 66 positive integers and found some cycles: 160>217>352>160 250>133>55>250 and I found these four that terminate themselves or any others if they appear (1 is trivial) 153 370 371 407 For the first 66 I got 18 "153"s so .2727... maybe approaching 1/3, but more suspiciously like 1/4... I looked around OEIS and saw this process as the base 10 third power version of the Narcissistic/Plus Perfect/Armstrong numbers... and found these four numbers to be the only ones for "n"=3 where an "n" digit number is the sum of the "n"th powers of its digits. Is there an analytical approach to determine the 1/3 claim for 153? 


#2
Feb213, 05:05 PM

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P: 11,589

Every number above 2915 will get smaller in the process, as 9999 > 2916 > 954 and numbers below 9999 will get smaller values, of course. Therefore, to find all possible final numbers or final series, it is sufficient to check the first 2915 numbers (probably less with a better analysis). For each n, you can consider the fraction of numbers below ending in 153. That fraction might have a limit, and it could be 1/3. But usually, there is no (known) way to derive that. I would not expect a proof  and even if there is one, I would not expect that it is understandable without a lot of number theory. For an example of a similar problem, see the Collatz conjecture  the problem is easy to understand, but hard to solve. 


#3
Feb313, 01:29 AM

P: 531

"Every number above 2915 will get smaller in the process..."
How did you determine that? 


#4
Feb313, 06:34 AM

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P: 11,589

Narcissistic/Plus Perfect/Armstrong
With the analysis after the quoted part.
In general, let f(n) be a function which maps a natural number to its sum of cubed digits. For any number n with d digits, the maximal f(n) corresponds to the number where all digits are 9. Therefore, f(n)<=9^3*d At the same time, any number with d digits is at least 10^(d1) (e.g. any number with 3 digits is at least 100). Therefore, n>=10^(d1). Using those equations, is easy to verify that f(n)<n for all numbers with more than 4 digits. For d=4, the analysis can be found in my previous post. 


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