A175975 Numbers k with the property that k^k has exactly two 1's.

8, 12, 17, 22, 23, 27, 30

Offset: 1

Author

Zak Seidov, Nov 02 2010

Keywords

• nonn
• base
• more
• less
• hard

Comments

Next term (if any) > 10000. Is the sequence finite?

Any subsequent terms are > 10^5. - Lucas A. Brown, Mar 20 2024

From David A. Corneth, Mar 20 2024: (Start)

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)

Examples

			8^8 = 16777216,
12^12 = 8916100448256,
17^17 = 827240261886336764177,
22^22 = 341427877364219557396646723584,
23^23 = 20880467999847912034355032910567,
27^27 = 443426488243037769948249630619149892803,
30^30 = 205891132094649000000000000000000000000000000.
		

Crossrefs

Cf. A000312 (n^n), A043494.

Programs

Mathematica

Select[Range[40],DigitCount[#^#,10,1]==2&] (* Harvey P. Dale, Dec 16 2013 *)

Python

A175975_list = [n for n in range(1000) if str(n**n).count('1') == 2] # Chai Wah Wu, May 19 2020