A245682 Numbers x whose digits can be permuted to produce more than a single multiple of x.

123876, 142857, 153846, 230769, 285714, 1028574, 1218753, 1238760, 1239876, 1246878, 1294857, 1402857, 1420785, 1425897, 1428507, 1428570, 1428597, 1428705, 1429857, 1485792, 1492857, 1538460, 1539846, 1570284, 1584297, 2300769, 2307690, 2307699, 2309769, 2857014, 2857140, 2859714, 2985714, 10028574, 10178649

Offset: 1

Author

Paolo P. Lava, Jul 29 2014

Keywords

• nonn
• base

Comments

It is a subset of A245680.

If x < 10^d is in the sequence, then so are x*10^j*(1+10^d+...+10^k*d) for all nonnegative integers j and k. - Robert Israel, Jul 29 2014

Examples

			Two permutations of 123876 are 371628, 867132  and  371628 / 123876 = 3, 867132  / 123876 = 7.
Five permutations of 142857 are 285714, 428571, 571428, 714285, 857142 and 285714 / 142857 = 2, 428571 / 142857 = 3, 571428 / 142857 = 4, 714285 / 142857 = 5, 857142 / 142857 = 6.
		

Links

• Paolo P. Lava, Table of n, a(n) for n = 1..200

Crossrefs

Cf. A008919, A096092, A096093, A245680.

Programs

Maple

P:=proc(q) local a,b,c,i,j,k,n,t; for n from 1 to q do a:=n; b:=[];
while a>0 do b:=[a mod 10,op(b)]; a:=trunc(a/10); od;
t:=0; for i from 2 to 9 do a:=i*n; c:=[];
while a>0 do c:=[a mod 10,op(c)]; a:=trunc(a/10); od;
if sort(b)=sort(c) then t:=t+1; fi; if t>1 then print(n); break;
fi; od; od; end: P(10^10);
# Alternative
N:= 10: # get a(1) to a(N)
count:= 0:
for x from 10 while count < N do
  M:= 10^(ilog10(x)+1)-1;
  L:= sort(convert(x,base,10));
  mults:= 0;
  for i from 2 to floor(M/x) do
    Lp:= sort(convert(i*x,base,10));
    if Lp = L then
      mults:= mults+1;
      if mults = 2 then
        count:= count+1;
        A[count]:= x;
        print(x);
        break;
      fi
    fi
   od
od:
seq(A[i],i=1..count); # Robert Israel, Jul 29 2014

PARI

for(n=1,10^8,d=vecsort(digits(n));p=0;for(k=2,9,dd=vecsort(digits(n*k));if(d==dd,p++));if(p>1,print1(n,", "))) \\ faster program Derek Orr, Jul 29 2014

Python

import itertools
from itertools import permutations
for n in range(1,10**8):
  plist = list(permutations(str(n)))
  count = 0
  lst = []
  for i in plist:
    num = ''
    for j in range(len(i)):
      num += i[j]
    if int(num)%n==0 and int(num)/n > 1:
      if int(num) not in lst:
        lst.append(int(num))
        count += 1
  if count > 1:
    print(n,end=', ') # Derek Orr, Jul 29 2014

Extensions

a(7) to a(10) from Robert Israel, Jul 29 2014

a(11) - a(35) from Derek Orr, Jul 29 2014