A016069 Numbers k such that k^2 contains exactly 2 distinct digits.

4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 20, 21, 22, 26, 30, 38, 88, 100, 109, 173, 200, 212, 235, 264, 300, 1000, 2000, 3000, 3114, 10000, 20000, 30000, 81619, 100000, 200000, 300000, 1000000, 2000000, 3000000, 10000000, 20000000, 30000000, 100000000, 200000000, 300000000

Offset: 1

Author

Robert G. Wilson v

Keywords

• nonn
• base
• nice

Comments

10^k, 2*10^k, 3*10^k for k > 0 are terms. - Chai Wah Wu, Dec 17 2021

Subsequence of primes is A057659. - Bernard Schott, Jul 29 2022

Examples

			26 is in the sequence because 26^2 = 676 contains exactly 2 distinct digits.
		

References

• R. K. Guy, Unsolved Problems in Number Theory, F24.

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..81
• Eric Weisstein's World of Mathematics, Square Number

Crossrefs

Cf. A016070, A018884, A018885, A057659.

Programs

Haskell

import Data.List (nub)
a016069 n = a016069_list !! (n-1)
a016069_list = filter ((== 2) . length . nub . show . (^ 2)) [0..]
-- Reinhard Zumkeller, Apr 14 2011

Magma

[n: n in [0..20000000] | #Set(Intseq(n^2)) eq 2]; // Vincenzo Librandi, Nov 04 2014

Mathematica

Join[Select[Range[90000],Count[DigitCount[#^2],?(#!=0&)]==2&],Flatten[ NestList[ 10#&,{100000,200000,300000},5]]] (* _Harvey P. Dale, Mar 09 2013 *)
Select[Range[20000000], Length[Union[IntegerDigits[#^2]]]==2&] (* Vincenzo Librandi, Nov 04 2014 *)

PARI

/* needs version >= 2.6 */
for (n=1, 10^9, if ( #Set(digits(n^2))==2, print1(n,", ") ) );
/* Joerg Arndt, Mar 09 2013 */

Python

from gmpy2 import is_square, isqrt
from itertools import combinations, product
A016069_list = []
for g in range(2,10):
    n = 2**g-1
    for x in combinations('0123456789',2):
        for i,y in enumerate(product(x,repeat=g)):
            if i > 0 and i < n and y[0] != '0':
                z = int(''.join(y))
                if is_square(z):
                    A016069_list.append(int(isqrt(z)))
A016069_list = sorted(A016069_list) # Chai Wah Wu, Nov 03 2014

Formula

a(n) = ((n-1) mod 3 + 1)*10^(ceiling(n/3)-7) for n >= 34 (conjectured). - Chai Wah Wu, Dec 17 2021