|
%I #57 Feb 10 2024 00:07:38
%S 2,3,7,8,22,47,53,77,92,157,187,188,192,552,558,577,707,772,922,2522,
%T 8338,17177,66888,575757,929522,1717177,8888588
%N None of the digits in k is present in k^2 or k^3.
%C If it exists, a(28) > 10^9. - _Derek Orr_, Jul 13 2014
%C a(28) > 2 * 10^9. - _Pratik Koirala_, _Eugene Fiorini_, _Otis Tweneboah_, _Nathan Fox_, Jul 13 2015
%C From _Manfred Scheucher_, Jul 23 2015: (Start)
%C I strongly conjecture that a(28) does not exist.
%C If a(28) exists, then a(28) > 10^20.
%C (End) [Edited by _Peter Munn_, Feb 08 2024 and _Manfred Scheucher_, Feb 09 2024 ]
%H Manfred Scheucher, <a href="/A029790/a029790_3.py.txt">corrected python script</a>
%e For k = 47, k^2 = 2209 and k^3 = 103823. 4 and 7 do not appear in either of these numbers.
%p filter:= proc(n) local S1,S23;
%p S1:= convert(convert(n,base,10),set);
%p S23:= convert(convert(n^2,base,10),set) union
%p convert(convert(n^3,base,10),set);
%p nops(S1 intersect S23)=0
%p end proc:
%p select(filter,[$1..10^5]); # _Robert Israel_, Jul 14 2014
%t Select[Range@ 1000000, Intersection[IntegerDigits[#^2], IntegerDigits@ #] == {} && Intersection[IntegerDigits[#^3], IntegerDigits@ #] == {} &] (* _Michael De Vlieger_, Jul 23 2015 *)
%o (Python)
%o def a(n):
%o s = str(n)
%o s2 = str(n**2)
%o s3 = str(n**3)
%o count = 0
%o for i in s:
%o if s2.count(i) == 0 and s3.count(i) == 0:
%o count += 1
%o else:
%o break
%o if count == len(s):
%o return True
%o n = 1
%o while n < 10**9:
%o if a(n):
%o print(n, end=', ')
%o n += 1
%o # _Derek Orr_, Jul 13 2014
%o (PARI) isok(n) = d = vecsort(Set(digits(n))); dd = vecsort(Set(digits(n^2))); ddd = vecsort(Set(digits(n^3))); for (i=1, #d, if (vecsearch(dd, d[i]) || vecsearch(ddd, d[i]), return (0));); 1 \\ _Michel Marcus_, Jul 13 2014
%o (PARI) is(n)=#setintersect(setunion(Set(digits(n^2)),Set(digits(n^3))), Set(digits(n)))==0 \\ _Charles R Greathouse IV_, Jul 23 2015
%K nonn,base,more,hard
%O 1,1
%A _Patrick De Geest_
|
