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Numbers n such that n and n^2 combined use different digits.
5

%I #29 Mar 23 2021 16:01:37

%S 2,3,4,7,8,9,17,18,24,29,53,54,57,59,72,79,84,209,259,567,807,854

%N Numbers n such that n and n^2 combined use different digits.

%C There are exactly 22 solutions in base 10.

%C More precisely: the concatenation of n and n^2 does not contain any digit twice. - _M. F. Hasler_, Oct 16 2018

%C a(20) = 567 and a(22) = 854 are the only two numbers k such that k and k^2 combined use each of the digits 1 to 9 exactly once (reference David Wells): 567^2 = 321489 and 854^2 = 729316. - _Bernard Schott_, Mar 23 2021

%D M. Kraitchik, Mathematical Recreations, p. 48, Problem 12. - From _N. J. A. Sloane_, Mar 15 2013

%D David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, 1997, page 144, entry 567.

%p M:=1000;

%p a1:=[]; a2:=[];

%p for n from 1 to M do

%p # are digits of n and n^2 distinct?

%p t1:=convert(n,base,10);

%p t2:=convert(n^2,base,10);

%p s3:={op(t1),op(t2)};

%p if nops(t1)+nops(t2) = nops(s3) then a1:=[op(a1),n]; a2:=[op(a2),n^2]; fi;

%p od:

%p a1; a2;

%p # _N. J. A. Sloane_, Mar 15 2013

%t Select[Range[10000], Intersection[IntegerDigits[ # ], IntegerDigits[ #^2]] == {} && Length[Union[IntegerDigits[ # ], IntegerDigits[ #^2]]] == Length[IntegerDigits[ # ]] + Length[IntegerDigits[ #^2]] &] (* _Tanya Khovanova_, Dec 25 2006 *)

%t Select[Range[10^3], Union@ Tally[Flatten@ IntegerDigits@ {#, #^2}][[All, -1]] == {1} &] (* _Michael De Vlieger_, Oct 17 2018 *)

%o (PARI) select( is(n)=#Set(Vecsmall(n=Str(n,n^2)))==#n, [1..999]) \\ _M. F. Hasler_, Oct 16 2018

%Y Cf. A059931, A029783 (digits of n are not present in n^2), A112736 (numbers whose squares are exclusionary).

%K nonn,base,fini,full

%O 1,1

%A _Patrick De Geest_, Feb 15 2001