|
|
A151752
|
|
a(n) is the unique n-digit number with all digits odd that is divisible by 5^n.
|
|
3
|
|
|
5, 75, 375, 9375, 59375, 359375, 3359375, 93359375, 193359375, 3193359375, 73193359375, 773193359375, 3773193359375, 73773193359375, 773773193359375, 5773773193359375, 15773773193359375, 515773773193359375, 7515773773193359375, 97515773773193359375
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Another way to phrase the proof of uniqueness: after we take the last n-1 digits to be the previous number in the sequence, all odd possibilities for the first digit give different remainders mod 5. By the pigeonhole principle, exactly one of them generates the required number. - Tanya Khovanova, Jun 18 2009
|
|
LINKS
|
|
|
FORMULA
|
a(n) = d(n)*10^(n-1) + a(n-1), where d(n), the leading digit of a(n), is one of the odd digits 1, 3, 5, 7, or 9 (forming the complete set of residues modulo 5) and is uniquely defined by the congruence: d(n) == (-a(n-1) / 10^(n-1)) (mod 5). - Max Alekseyev
|
|
MAPLE
|
a:= proc(n) option remember; local k, l;
if n=1 then 5
else l:= a(n-1);
for k from 1 to 9 by 2
while (parse(cat(k, l)) mod 5^n)<>0 do od;
parse(cat(k, l))
fi
end:
|
|
MATHEMATICA
|
nxt[n_]:=Module[{x=FromDigits/@(Prepend[IntegerDigits[n], # ]&/@{1, 3, 5, 7, 9}), l}, l=IntegerLength[n]+1; First[Select[x, Mod[ #, 5^l]==0&]]]; NestList[nxt, 5, 25] (* Harvey P. Dale, Jul 06 2009 *)
|
|
PROG
|
(Magma) v:=[5];
for i in [2..20] do
for s in [1, 3, 5, 7, 9] do
v[i]:=s*10^(i-1)+v[i-1];
if v[i] mod 5^i eq 0 then
break;
end if;
end for;
end for;
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|