This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(4) = 23 + 27 + 29 + 34 = 113 is prime.
35 is included because 5 is a divisor of 35, but 37 is not included because neither 3 nor 7 is a divisor of 37.
a038770 n = a038770_list !! (n-1)
a038770_list = filter f [1..] where
f u = g u where
g v = v > 0 && (((d == 0 || r > 0) && g v') || r == 0)
where (v',d) = divMod v 10; r = mod u d
-- Reinhard Zumkeller, Jul 30 2015, Jun 19 2011
Select[Range[120],MemberQ[Divisible[#,Cases[IntegerDigits[#],Except[0]]], True]&] (* Harvey P. Dale, Jun 20 2011 *) Select[Range[120],AnyTrue[#/DeleteCases[IntegerDigits[#],0],IntegerQ]&] (* Harvey P. Dale, Mar 29 2024 *)
is(n)=my(v=vecsort(eval(Vec(Str(n))),,8));for(i=if(v[1],1,2),#v,if(n%v[i]==0,return(1)));0 \\ Charles R Greathouse IV, Jul 22 2011
def ok(n): return any(n%int(d) == 0 for d in str(n) if d != '0') print(list(filter(ok, range(1, 103)))) # Michael S. Branicky, May 20 2021
a(35)=1 because 5 is a divisor of 35 but 3 is not.
import Data.Char (digitToInt)
a038769 n = length $ filter (== 0)
$ map ((mod n) . digitToInt) $ filter (> '0') $ show n
-- Reinhard Zumkeller, Jun 19 2011
[#[c:c in Intseq(k) |not IsZero(c) and k mod c eq 0]:k in [1..105]]; // Marius A. Burtea, Dec 23 2019
f:= proc(n) local L; L:= convert(n,base,10); nops(select(t -> t > 0 and n mod t = 0, L)) end proc: map(f, [$1..1000]); # Robert Israel, Jul 04 2016
Array[Count[Position[Most@ DigitCount@ #, ?(# > 0 &)][[All, 1]], k /; Mod[#, k] == 0] &, 105] (* Michael De Vlieger, Dec 23 2019 *) Table[Count[n/Select[IntegerDigits[n],#>0&],?IntegerQ],{n,110}] (* _Harvey P. Dale, Mar 04 2023 *)
9 is excluded because it can be written as 111111111, 1001, 100, 21, 14, 13, 12, 11, 10 or 9 and in every case there is a digit which divides 9; 11 is in the sequence because in base 4 it is written 23 and 11 is not divisible by either 2 or 3
23 is the first number that is not divisible by either of its digits. 37 and 38 are the first two consecutive numbers that are not divisible by any of their digits. Thus, a(2) = 37. 56, 57, 58 (and 59) are the first three (and four) consecutive numbers that are not divisible by any of their digits. Thus, a(3) = a(4) = 56. 866, 867, 868, 869, and 870 are the first five consecutive numbers that are not divisible by any of their digits. Thus, a(5) = 866.
def DivDig(x):
total = 0
for i in str(x):
if i != '0':
if x/int(i) % 1 == 0:
return True
return False
def Nums(x):
n = 1
while n < 10**3:
count = 0
for i in range(n,n+x):
if not DivDig(i):
count += 1
else:
break
if count == x:
return n
else:
n += 1
x = 1
while x < 10:
print(Nums(x))
x += 1
2809 is in the sequence because 2809 is not divisible by 2, 8 or 9.
Select[Range[350]^2,NoneTrue[#/(IntegerDigits[#]/.(0->Nothing)), IntegerQ]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Oct 08 2017 *)
isOK(n) = my(v=vecsort(digits(n^2), , 8)); for(i=1+(v[1]==0), #v, if(n^2%v[i]==0, return(0))); 1 s=[]; for(n=1, 1000, if(isOK(n), s=concat(s, n^2))); s
50653 is in the sequence because 37^3 = 50653 is not divisible by 3, 5 or 6.
ndnzdQ[n_]:=NoneTrue[n/Select[IntegerDigits[n],#!=0&],IntegerQ]; Select[ Range[ 400]^3,ndnzdQ] (* Harvey P. Dale, Jan 16 2022 *)
isOK(n) = my(v=vecsort(digits(n^3), , 8)); for(i=1+(v[1]==0), #v, if(n^3%v[i]==0, return(0))); 1 s=[]; for(n=1, 1000, if(isOK(n), s=concat(s, n^3))); s
a(1) = 17 and 2*17 = 34 is not divisible by 3 or 4; a(2) = 19 and 2*19 = 38 is not divisible by 3 or 8; a(3) = 23 and 2*23 = 46 is not divisible by 4 or 6; a(4) = 27 and 2*27 = 54 is not divisible by 5 or 4; etc. 31 is not in the sequence as 2*31 = 62 is divisible by 2.
q[n_] := AllTrue[IntegerDigits[2*n], # == 0 || !Divisible[2*n, #] &]; Select[Range[350], q] (* Amiram Eldar, Mar 14 2022 *) nnzdQ[n_]:=NoneTrue[Mod[2 n,Select[IntegerDigits[2 n],#!=0&]],#==0&]; Select[Range[350],nnzdQ] (* Harvey P. Dale, Apr 05 2025 *)
def ok(n): return not any(2*n%int(d)==0 for d in set(str(2*n)) if d!='0') print([k for k in range(1, 344) if ok(k)]) # Michael S. Branicky, Mar 14 2022
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