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
Alois P. Heinz, Table of n, a(n) for n = 1..300
33rd USAMO 2003, Problem 1
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:
seq(a(n), n=1..30); # Alois P. Heinz, Jun 18 2009
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;
v; // Marius A. Burtea, Mar 18 2019
CROSSREFS
KEYWORD
nonn,base
AUTHOR
David W. Wilson, Jun 16 2009
EXTENSIONS
More terms from Max Alekseyev, Jun 17 2009
Further terms from Alois P. Heinz, Jun 18 2009
More terms from Harvey P. Dale, Jul 06 2009
STATUS
approved