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A331492
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Numbers k such that the digits of k^(1/5) begin with k.
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6
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0, 1, 17, 315, 316, 5623, 99999, 100000, 1778279, 31622776, 562341324, 562341325, 9999999999, 10000000000, 177827941003, 3162277660168, 56234132519034, 999999999999999, 1000000000000000, 17782794100389227, 17782794100389228, 316227766016837932, 316227766016837933
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OFFSET
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1,3
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COMMENTS
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The following algorithm will generate all numbers k such that the digits of k^(1/b) begins with k: For each integer m >= 0, compute r = floor(10^(bm/(b-1)). Let s <= r be the largest integer >= 0 such that (r-s)*10^(bm) < (r-s+1)^b. Then r, r-1, ... r-s are such numbers k and there are no other such numbers.
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LINKS
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EXAMPLE
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5623^(1/5) = 5.6233305990931... which starts with the digits 5623, so 5623 is in the sequence.
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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