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A240387
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Least k such that sum of n-th power of digits of k equals sum of prime divisors of k.
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0
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2, 12, 142, 210, 22011, 210120, 101220, 11012210, 2202120, 11210021221, 1100200012, 1000200000111, 1211201222, 211222020, 101101001120
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OFFSET
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1,1
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COMMENTS
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LINKS
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EXAMPLE
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a(5) = 22011 = 3*11*23*29 because 2^5 + 2^5 + 0^5 + 1^5 + 1^5 = 3 + 11 + 23 + 29 = 66.
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MATHEMATICA
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Do[k=2; While[!Total[Transpose[FactorInteger[k]][[1]]]==Total[IntegerDigits[k]^n], k++]; Print[n, " ", k], {n, 1, 15}]
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CROSSREFS
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KEYWORD
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nonn,base,hard
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AUTHOR
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STATUS
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approved
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