OFFSET
1,2
COMMENTS
Note that the values of n for which a(n) = 1 have density 1.
Is it known that a(n)=0 only for n a power of 10? - Christopher J. Smyth, Aug 21 2014
a(n) >= ceiling(log_n(10)*9), whenever a(n)>0. This is because in order for an integer to have 10 digits its base-10 magnitude must be at least 9. - Ely Golden, Sep 06 2017
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
Ely Golden, Python program to generate a(n)
FORMULA
a(10^e) = 0; a(m^e) = a(m)/e for e dividing a(m). - Reinhard Zumkeller, Dec 06 2004
EXAMPLE
a(5)=19: 5^19 = 19073486328125.
MAPLE
a:= proc(n) local k;
if n = 10^ilog10(n) then return 0 fi;
for k from 1 do
if nops(convert(convert(n^k, base, 10), set))=10 then return k fi
od
end proc:
seq(a(n), n=1..100); # Robert Israel, Aug 20 2014
MATHEMATICA
Table[If[IntegerQ@ Log10[n], 0, SelectFirst[Range[#, # + 100] &@ Ceiling[9 Log[n, 10]], NoneTrue[DigitCount[n^#], # == 0 &] &]], {n, 71}] (* Michael De Vlieger, Sep 06 2017 *)
PROG
(PARI) a(n) = if (n == 10^valuation(n, 10), return (0)); k=1; while(#vecsort(digits(n^k), , 8)!=10, k++); k; \\ Michel Marcus, Aug 20 2014
(Python)
def a(n):
s = str(n)
if n == 1 or (s.count('0')==len(s)-1 and s.startswith('1')):
return 0
k = 1
count = 0
while count != 10:
count = 0
for i in range(10):
if str(n**k).count(str(i)) == 0:
count += 1
break
if count:
k += 1
else:
return k
n = 1
while n < 100:
print(a(n), end=', ')
n += 1
# Derek Orr, Aug 20 2014
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Amarnath Murthy, Dec 03 2003
EXTENSIONS
More terms from Reinhard Zumkeller, Dec 06 2004
Corrected a(15), a(17), a(38), a(48), a(56) and a(65). (For each of these terms, the only 1 in n^k is the first digit.) - Jon E. Schoenfield, Sep 20 2008
STATUS
approved