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A034709
Numbers divisible by their last digit.
35
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, 111, 112, 115, 121, 122, 123, 124, 125, 126, 128, 131, 132
OFFSET
1,2
COMMENTS
The differences between consecutive terms repeat with period 1177 and the corresponding terms differ by 2520 = LCM(1,2,...,9). In other words, a(k*1177+i) = 2520*k + a(i). - Giovanni Resta, Aug 20 2015
The asymptotic density of this sequence is 1177/2520 = 0.467063... (see A341431 and A341432 for the values in other base representations). - Amiram Eldar, Nov 24 2022
LINKS
MAPLE
N:= 1000: # to get all terms <= N
sort([seq(seq(ilcm(10, d)*x+d, x=0..floor((N-d)/ilcm(10, d))), d=1..9)]); # Robert Israel, Aug 20 2015
MATHEMATICA
dldQ[n_]:=Module[{idn=IntegerDigits[n], last1}, last1=Last[idn]; last1!= 0&&Divisible[n, last1]]; Select[Range[150], dldQ] (* Harvey P. Dale, Apr 25 2011 *)
Select[Range[150], Mod[#, 10]!=0&&Divisible[#, Mod[#, 10]]&] (* Harvey P. Dale, Aug 07 2022 *)
PROG
(Haskell)
import Data.Char (digitToInt)
a034709 n = a034709_list !! (n-1)
a034709_list =
filter (\i -> i `mod` 10 > 0 && i `mod` (i `mod` 10) == 0) [1..]
-- Reinhard Zumkeller, Jun 19 2011
(Python)
A034709_list = [n for n in range(1, 1000) if n % 10 and not n % (n % 10)]
# Chai Wah Wu, Sep 18 2014
(PARI) for(n=1, 200, if(n%10, if(!(n%digits(n)[#Str(n)]), print1(n, ", ")))) \\ Derek Orr, Sep 19 2014
KEYWORD
nonn,base,easy,nice
STATUS
approved