OFFSET
1,3
COMMENTS
The number 82000 is famous for having only digits 0 and 1 in all bases <= 5, no other such number > 1 is known. See also A146025 and A258981.
If explicit formulas for (convenient) infinite subsequences of this one can be found, this could open new ways to progress on this problem.
The terms come in groups having roughly the first half (or at least third) of digits in common, see the link "Terms in base 10, 5 and 3".
LINKS
Ray Chandler, Table of n, a(n) for n = 1..10000 (first 84 terms from M. F. Hasler, next 31 terms from Charles R Greathouse IV)
M. F. Hasler, Terms in base 10, 5 and 3.
FORMULA
a(n) >> n^k with k = log 5/log 2 = 2.321928.... - Charles R Greathouse IV, Feb 02 2016
MAPLE
d:= 20: # to get all terms < 5^d
res:= NULL:
T:= combinat:-cartprod([[$0..1]$d]):
while not T[finished] do
r:= T[nextvalue]();
v:= add(r[i]*5^(d-i), i=1..d);
if max(convert(v, base, 3)) <= 1 then
res:= res, v
fi
od:
res; # Robert Israel, Feb 01 2016
MATHEMATICA
Module[{t=Tuples[{0, 1}, 25], b3, b5}, b3=FromDigits[#, 3]&/@t; b5=FromDigits[ #, 5]&/@t; Intersection[b3, b5]] (* The program generates the first 26 terms of the sequence. *) (* Harvey P. Dale, Dec 13 2021 *)
PROG
(PARI) print1(0); for(n=1, 1e10, vecmax(digits(t=subst(Pol(binary(n)), 'x, 5), 3))<2&&print1(", "t))
(PARI) list(lim)=my(v=List([0]), d=digits(lim\1, 5), t); for(i=1, #d, if(d[i]>1, for(j=i, #d, d[j]=1); break)); for(n=1, fromdigits(d, 5), t=fromdigits(binary(n), 5); if(vecmax(digits(t, 3))<2, listput(v, t))); Vec(v) \\ Charles R Greathouse IV, Feb 02 2016
CROSSREFS
KEYWORD
nonn,base
AUTHOR
M. F. Hasler, Feb 01 2016
STATUS
approved