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.
Select[Range[120],ContainsAny[IntegerDigits[#],{4,6,8,9}]&] (* Harvey P. Dale, Sep 06 2018 *)
125 is in the sequence because the proper divisors of 125 are {1, 5, 25} with the same digits as 125.
Pn[n_]:=Sort[DeleteDuplicates[Flatten[IntegerDigits[Take[Divisors[n],DivisorSigma[0,n]-1]]]]];Fn[n_]:=Sort[DeleteDuplicates[IntegerDigits[n]]];lst={};Do[If[!PrimeQ[n]&&Pn[n]===Fn[n],AppendTo[lst,n]],{n,2,10^5}];lst
Select[Range[10^5], ! PrimeQ@# && Union@ IntegerDigits@ # == Union @@ IntegerDigits /@ Most@ Divisors@ # &] (* Giovanni Resta, Feb 12 2014 *)
isok(n) = { my(digs = []); fordiv(n, d, if (d != n, digs = concat(digs, digits(d)))); (n != 1) && !isprime(n) && vecsort(digs,,8) == vecsort(digits(n),,8);} \\ Michel Marcus, Feb 12 2014
5^1 = 5 does not contain a 1 but 5^2 = 25 does contain a 2 so a(5) = 2. 7^1 = 7 does not contain a 1, 7^2 = 49 does not contain a 2, but 7^3 = 343 does contain a 3 so a(7) = 3.
f:= proc(n) local k;
for k from 1 to 9 do
if member(k,convert(n^k,base,10)) then return k fi
od;
FAIL
end proc:
map(f, [$1..100]); # Robert Israel, Sep 16 2024
a[n_] := Block[{k=1}, While[{} == StringPosition[ ToString[n^k], ToString[k]], k++]; k]; Array[a, 84] (* Giovanni Resta, Mar 11 2014 *)
sk[n_]:=Module[{k=1},While[SequenceCount[IntegerDigits[n^k],IntegerDigits[k]] == 0,k++];k]; Array[sk,90] (* Harvey P. Dale, May 12 2022 *)
def Sub(x):
for n in range(10**3):
if str(x**n).find(str(n)) > -1:
return n
x = 1
while x < 10**3:
print(Sub(x))
x += 1
a(4) = 22 since it contains the largest digit from a(2) = 2, and can't be 12 since the digit 1 appears in a(3) = 10.
Nest[Append[#, Block[{k = 2, d}, While[Nand[FreeQ[#[[All, 1]], k], MemberQ[Set[d, IntegerDigits[k]], Max[#[[-2, -1]] ] ], ! IntersectingQ[d, #[[-1, -1]]] ], k++]; {k, d}]] &, Transpose@ {#, IntegerDigits@ #} &@ Range[2], 73][[All, 1]] (* Michael De Vlieger, Apr 12 2018 *)
\\ See Links section.
The multiples of 3 are: 3, 6, 9, 12, 15, etc.; 12 is the first one containing the digit 1, hence a(3) = 12.
on1[n_]:=Module[{k=1},While[DigitCount[k*n,10,1]<1,k++];k*n]; Array[on1,60] (* Harvey P. Dale, Apr 09 2022 *)
a(n) = forstep (m=n, oo, n, if (setsearch(Set(digits(m)), 1), return (m)))
16 is semi-1 because 1,10,11,12,13,14,15,16 have a 1 in them, there are 8 such numbers, and 8 is half of 16. 2 is semi-1 because 1 has a 1 in it and 2 does not.
s = {}; c = 0; Do[If[DigitCount[n, 10, 1] > 0, c++]; If[n == 2*c, AppendTo[s, n]], {n, 1, 1062880}]; s (* Amiram Eldar, May 01 2021 *)
With[{nn=11*10^5},Select[Partition[Riffle[Range[nn],Accumulate[Table[If[DigitCount[n,10,1]>0,1,0],{n,nn}]]],2],#[[1]]==2#[[2]]&]][[;;,1]] (* Harvey P. Dale, Jun 23 2023 *)
lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, #va, va[n] = va[n-1] + (#select(x->(x==1), digits(n)) > 0);); for (n=1, nn, if (va[n] == n/2 , print1(n, ", ")););} \\ Michel Marcus, May 02 2021
for (1..10000000) {
if (/1/) {
$s++;
}
if ($_==2*$s) {
print $_."\n";
}
}
The proper divisors of 54 are 1, 2, 3, 6, 9, 18 and 27; none of them appear in the decimal expansion of 54 so 54 belongs to this sequence.
A383749Q[k_] := SelectFirst[Divisors[k], StringContainsQ[IntegerString[k], IntegerString[#]] &] == k; Select[Range[500], A383749Q] (* Paolo Xausa, May 12 2025 *)
is(n, base = 10) = {
my (d = digits(n, base));
for (i = 1, #d,
if (d[i],
for (j = i, #d,
if ((i!=1 || j!=#d) && n % fromdigits(d[i..j], base)==0,
return (0);););););
return (1);}
from itertools import count, islice from sympy import divisors def A383749_gen(startvalue=1): # generator of terms >= startvalue return filter(lambda n:not any(dA383749_list = list(islice(A383749_gen(),40)) # Chai Wah Wu, May 10 2025
def ok(n):
s = str(n)
subs = (s[i:j] for i in range(len(s)) for j in range(i+1, len(s)+1) if s[i]!='0')
return n and not any(n%v == 0 for ss in subs if n > (v:=int(ss)) > 0)
print([k for k in range(308) if ok(k)]) # Michael S. Branicky, May 11 2025
Select[Range[200], ! PrimeQ[#] && MemberQ[IntegerDigits[#], 1] &] (* T. D. Noe, Mar 06 2012 *)
The blocks of terms including at least a digit "1" are indicated here by parentheses; the successive block-sizes are 1, 2, 3, 4, 5,... which reproduces the sequence itself: (1),2,3,4,5,6,7,8,9,(10,11),20,(12,13,14),22,(15,16,17,18),23,(19,21,31,41,51),24...
isA011531 := proc(n) nops(convert(convert(n,base,10),set) intersect {1}) > 0 ; simplify(%) ; end proc: A011531_next := proc(n) local a; for a from n+1 do if isA011531(a) then return a; end if; end do: end proc: isA052383 := proc(n) not isA011531(n) ; end proc: A052383_next := proc(n) local a; for a from n+1 do if isA052383(a) then return a; end if; end do: end proc: A275512grp1 := proc(a) local idx1,n ; idx1 := 0 ; if isA052383(a) then return 0; end if; for n from 1 do if isA011531(A275512(n)) then if n =1 then idx1 := 1; elif isA052383(A275512(n-1)) then idx1 := idx1+1 ; end if; end if; if A275512(n) = a then return idx1 ; end if; end do: end proc: A275512 := proc(n) option remember; local a,aprev,prev1,grp1,d,seen,reqlen,npr; if n =1 then 1; else for a from 2 do seen := false; for npr from 1 to n-1 do if procname(npr) = a then seen := true; break; end if; end do: if not seen then aprev := procname(n-1) ; if isA052383(aprev) then return a; else prev1 := 0 ; for d from 1 to n-1 do if isA011531(procname(n-d)) then prev1 := prev1+1 ; else break; end if; end do: grp1 := A275512grp1(aprev) ; reqlen := procname(grp1) ; if reqlen > prev1 then return A011531_next(aprev) ; elif n-prev1-1 > 0 then return A052383_next(procname(n-prev1-1)) ; else return 2 ; end if; end if; end if; end do; end if; end proc: seq(A275512(n),n=1..100) ; # R. J. Mathar, Jul 31 2016
16325 is in the sequence because the set of the digits is E = {1, 2, 3, 5, 6} and the proper divisors (or aliquot parts) of 16325 are 1, 5, 25, 653 and 3265 with the same set of digits.
with(numtheory):
for n from 1 to 200000 do:
z:=convert(n,base,10):n0:=nops(z):lst1:={op(z),z[n0]}:
x:=divisors(n):n1:=nops(x):lst:={}:
for m from 1 to n1-1 do:
y:=convert(x[m],base,10):n2:=nops(y):
lst2:={op(y),y[n2]}:lst:=lst union lst2
od:
if lst1=lst then
printf(`%d, `,n):
else
fi:
od:
Select[Range[10^5], Function[k, Union@ Flatten@ Map[IntegerDigits, Most@ Divisors@ k] == Union@ IntegerDigits@ k]] (* Michael De Vlieger, Feb 25 2017 *)
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