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Topic: fibonacci iteration (Read 1164 times) |
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Christine
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fibonacci iteration
« on: Mar 26th, 2014, 10:12am » |
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If you take any Fibonacci number (except 1, 13) find the sum of the squares of its digits, and then repeat; you will come to 89 why is that?
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towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
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Re: fibonacci iteration
« Reply #1 on: Mar 26th, 2014, 10:39am » |
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It's true for 856929 of the first million non-negative integers, so it's not special to Fibonacci numbers. Also, it's not true for quite a few Fibonacci numbers, e.g.: 17 : 2584 18 : 4181 24 : 75025 31 : 2178309 32 : 3524578 44 : 1134903170 49 : 12586269025 82 : 99194853094755497 83 : 160500643816367088 86 : 679891637638612258 92 : 12200160415121876738 105 : 6356306993006846248183 108 : 26925748508234281076009 114 : 483162952612010163284885 116 : 1264937032042997393488322 121 : 14028366653498915298923761 125 : 96151855463018422468774568 129 : 659034621587630041982498215 131 : 1725375039079340637797070384 132 : 2791715456571051233611642553 133 : 4517090495650391871408712937 142 : 343358302784187294870275058337 In fact, it's about the same fraction as for regular integers, again pointing to Fibonacci numbers not being special.
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« Last Edit: Mar 26th, 2014, 10:57am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Christine
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Posts: 159
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Re: fibonacci iteration
« Reply #2 on: Mar 26th, 2014, 11:10am » |
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Thanks. I've just looked at prime curios! page http://primes.utm.edu/curios/page.php/89.html One entry, it reads: Hugo Steinhaus (1887-1972) proved that if you take any positive integer, find the sum of the squares of its digits, and then repeat; that you will either come to 1 or the cyclic sequence 145, 42, 20, 4, 16, 37, 58, 89. It doesn't mention about percentage of getting a 89 or any percentages of these numbers
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Christine
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Posts: 159
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Re: fibonacci iteration
« Reply #4 on: Mar 30th, 2014, 1:23pm » |
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thanks
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