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A029790
None of the digits in k is present in k^2 or k^3.
6
2, 3, 7, 8, 22, 47, 53, 77, 92, 157, 187, 188, 192, 552, 558, 577, 707, 772, 922, 2522, 8338, 17177, 66888, 575757, 929522, 1717177, 8888588
OFFSET
1,1
COMMENTS
If it exists, a(28) > 10^9. - Derek Orr, Jul 13 2014
a(28) > 2 * 10^9. - Pratik Koirala, Eugene Fiorini, Otis Tweneboah, Nathan Fox, Jul 13 2015
From Manfred Scheucher, Jul 23 2015: (Start)
I strongly conjecture that a(28) does not exist.
If a(28) exists, then a(28) > 10^20.
(End) [Edited by Peter Munn, Feb 08 2024 and Manfred Scheucher, Feb 09 2024]
If it exists, a(28) > 10^37. - Michael S. Branicky, May 05 2024
EXAMPLE
For k = 47, k^2 = 2209 and k^3 = 103823. 4 and 7 do not appear in either of these numbers.
MAPLE
filter:= proc(n) local S1, S23;
S1:= convert(convert(n, base, 10), set);
S23:= convert(convert(n^2, base, 10), set) union
convert(convert(n^3, base, 10), set);
nops(S1 intersect S23)=0
end proc:
select(filter, [$1..10^5]); # Robert Israel, Jul 14 2014
MATHEMATICA
Select[Range@ 1000000, Intersection[IntegerDigits[#^2], IntegerDigits@ #] == {} && Intersection[IntegerDigits[#^3], IntegerDigits@ #] == {} &] (* Michael De Vlieger, Jul 23 2015 *)
PROG
(Python)
def a(n):
s = str(n)
s2 = str(n**2)
s3 = str(n**3)
count = 0
for i in s:
if s2.count(i) == 0 and s3.count(i) == 0:
count += 1
else:
break
if count == len(s):
return True
n = 1
while n < 10**9:
if a(n):
print(n, end=', ')
n += 1
# Derek Orr, Jul 13 2014
(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
(PARI) is(n)=#setintersect(setunion(Set(digits(n^2)), Set(digits(n^3))), Set(digits(n)))==0 \\ Charles R Greathouse IV, Jul 23 2015
CROSSREFS
Sequence in context: A024540 A029785 A045545 * A369350 A206725 A129645
KEYWORD
nonn,base,more,hard
STATUS
approved