A178158 Numbers n that are divisible by every suffix of n.
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 21, 22, 24, 25, 31, 32, 33, 35, 36, 41, 42, 44, 45, 48, 51, 52, 55, 61, 62, 63, 64, 65, 66, 71, 72, 75, 77, 81, 82, 84, 85, 88, 91, 92, 93, 95, 96, 99, 101, 102, 104, 105, 125, 201, 202, 204, 205, 208, 225, 301, 302, 303, 304, 305, 306, 312, 315, 325, 375, 401, 402, 404, 405, 408, 425, 501, 502, 504, 505, 525, 601, 602, 603, 604
Offset: 1
Examples
9375 is in the sequence because : . 5 | 9375 ; . 75 | 9375 ; . 375 | 9375 ; . 9375 | 9375 .
Links
- Robert Israel, Table of n, a(n) for n = 1..10000 (first 788 terms from Reinhard Zumkeller)
- Math Overflow, The number of numbers no greater than n that are divisible by all their suffixes
Programs
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Haskell
import Data.List (tails) a178158 n = a178158_list !! (n-1) a178158_list = filter (\suff -> all ((== 0) . (mod suff)) (map read $ tail $ init $ tails $ show suff :: [Integer])) a067251_list -- Reinhard Zumkeller, Mar 26 2012
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Maple
with(numtheory):T:=array(1..5):for n from 1 to 10000 do:ind:=0:l:=length(n):n0:=n:s:=0:for m from 0 to l-1 do:q:=n0:u:=irem(q, 10):v:=iquo(q, 10):n0:=v : s:=s + u*10 ^m:if irem(n,10)<>0 and irem(n, s)=0 then ind:=ind+1:else fi:od:if ind=l then printf(`%d,`, n):else fi:od: # Alternative: filter:= proc(x) if x mod 10 = 0 then return false fi; andmap(t -> type(x/(x mod 10^t),integer), [$1..ilog10(x)]) end proc: Res:= $1..9: for d from 1 to 6 do for y from 1 to 9 do for z in sort(convert(select(`<`,numtheory:-divisors(y*10^d),10^d),list)) do if filter(y*10^d+z) then Res:= Res, y*10^d+z; fi od od od: Res; # Robert Israel, Oct 17 2018
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