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A175975
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Numbers k with the property that k^k has exactly two 1's.
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0
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OFFSET
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1,1
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COMMENTS
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Next term (if any) > 10000. Is the sequence finite?
Any subsequent terms are > 10^9.
One could assume the digits to be random and look at the last few digits of k^k until it is clear that the number of 1's exceeds two. That way there is no need to compute k^k entirely.
Powers of 10 have exactly one digit 1 and so cannot be terms.
For 401113652 we have the last 253 digits containing exactly two 1's while the last 254 digits contain three 1's, proving that 401113652 is not a term. It appears that it usually takes the last 20 digits to prove that a number is not a term (the mode value); the median appears to be 27 digits. (End)
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LINKS
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EXAMPLE
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8^8 = 16777216,
12^12 = 8916100448256,
17^17 = 827240261886336764177,
22^22 = 341427877364219557396646723584,
23^23 = 20880467999847912034355032910567,
27^27 = 443426488243037769948249630619149892803,
30^30 = 205891132094649000000000000000000000000000000.
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MATHEMATICA
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Select[Range[40], DigitCount[#^#, 10, 1]==2&] (* Harvey P. Dale, Dec 16 2013 *)
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PROG
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(Python)
A175975_list = [n for n in range(1000) if str(n**n).count('1') == 2] # Chai Wah Wu, May 19 2020
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CROSSREFS
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KEYWORD
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nonn,base,more,less,hard
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AUTHOR
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STATUS
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approved
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