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Topic: digit powers (Read 1315 times) |
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Christine
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digit powers
« on: Feb 22nd, 2012, 10:24am » |
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Number expressed as sum of integers raised to the power of its own digits:
For example,
10 = 9^1 + 9^0
3909511=5^3+5^9+5^0+5^9+5^5+5^1+5^1
4624= 4^4 + 4^6 + 4^2 + 4^4
with 10, the "base" integer is 9, with 3909511 it's 5, and 4624 it's 4.
How to find others?
Is the set of these numbers finite or infinite?
Can other integers 2,3,6,7,8,10... be the bases?
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towr
wu::riddles Moderator
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Re: digit powers
« Reply #1 on: Feb 22nd, 2012, 12:49pm » |
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here's some I found with a brute force search:
base, sum
9 10
3 12
37 111
8 1033
32 2112
4 4624
25 31301
I think in general it also holds for 10(2k+1)*9-1 with base [10(2k+1)*9-1]/[(2k+1)*9], but it's late and I'm not going to try and prove it.
If not for that sequence, it holds for 109^k-1 with base [109^k-1]/9k. So at least we can say there's an infinite number of them.
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Wikipedia, Google, Mathworld, Integer sequence DB
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