|
|
A055471
|
|
Divisible by the product of its nonzero digits.
|
|
9
|
|
|
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 20, 24, 30, 36, 40, 50, 60, 70, 80, 90, 100, 101, 102, 104, 105, 110, 111, 112, 115, 120, 128, 132, 135, 140, 144, 150, 175, 200, 208, 210, 212, 216, 220, 224, 240, 250, 300, 306, 312, 315, 360, 384, 400, 432, 480, 500
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
If n is the term then 10n also is. - Zak Seidov, Jun 09 2013
De Koninck and Luca showed that the number of terms of this sequence below x is at least x^0.495 but at most x^0.901 for sufficiently large x. - Tomohiro Yamada, Nov 18 2017
This sequence begins with a run of 12 consecutive terms, from 1 to 12. The maximal length of a run of consecutive integer terms is 13. The smallest example of such a run begins with 1111011111000 and ends with 1111011111012 (Diophante link). - Bernard Schott, Apr 26 2019
These numbers are called "nombres prodigieux" on the French site Diophante. - Bernard Schott, Apr 26 2019
|
|
LINKS
|
|
|
MATHEMATICA
|
Select[Range[5000], IntegerQ[ #/(Times @@ Select[IntegerDigits[ # ], # > 0 &])] &] (* Alonso del Arte, Aug 04 2004 *)
|
|
PROG
|
(MATLAB) m=1;
for n=1:1000
v=dec2base(n, 10)-'0';
v = v(v~=0);
if mod(n, prod(v))==0
sol(m)=n;
m=m+1;
end
end
(Magma) m:=1; sol:=[];
for n in [1..1000] do
v:=Intseq(n, 10);
while &*v eq 0 do; Exclude(~v, 0); end while;
if n mod &*(v) eq 0 then ; sol[m]:=n; m:=m+1; end if;
end for;
(Python)
from math import prod
def ok(n): return n > 0 and n%prod([int(d) for d in str(n) if d!='0']) == 0
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|