If I haven't made a mistake these 31 integers are the only "reduced" loopers: 25, 26, 28, 34, 35, 36, 38, 43, 46, 52, 53, 62, 63, 64, 82, 83, 236, 239, 246, 254, 296, 326, 329, 362, 392, 426, 462, 524, 542, 926, 962 By reduced I mean they don't contain 1's. As you pointed out one can insert as many 1's as desired in any of these and still have a looper. This sequence is not in the OEIS. On Tue, Dec 29, 2020 at 8:06 PM Allan Wechsler <acwacw@gmail.com> wrote:
Since the vast majority of integers have a 0 digit, the vast majority of integers stop on the first step.
Of the remainder, the vast majority will have a 1 digit, and so will enter a chain in which all the 1's get removed. Then, the vast majority will have a 2, and their lengths will double, and again the vast majority will have a 0 and stop.
So I expect loops to be rare and get rarer. The stoppers dominate so aggressively that I expect that there will be either a finite set of loopers, or at best a small number of easily-recognized infinite classes of loopers, with a linear or quadratic upper bound on the number of loopers in each order of magnitude.
25 is the smallest looper. I would be interested to see any example of a five-digit looper. (Essentially I am hoping that somebody writes the code for me.)
On Tue, Dec 29, 2020 at 12:20 PM Éric Angelini <bk263401@skynet.be> wrote:
Hello Math-Fun,
Say D has 5 digits: D = abcde Starting from the right, pick D’s smallest digit (say it is b) Then E = acde^b Repeat. Examples: 23 --> 3^2 = 9 = stop 34 --> 4^3 = 64 --> 6^4 = 1296 --> 296^1 = 296 --> 96^2 = 9216 --> 926^1 = 926 --> 96^2 = 9216 = loop 35 --> 5^3 = 125 --> 25^1 = 25 --> 5^2 = 25 = loop 3252 --> 325^2 = 105625 --> 15625^0 = 1 = stop Which integers loop? Which integers stop? Best, É. (sorry if this has been explored before)
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