A023086 Numbers k such that k and 2*k are anagrams.

0, 125874, 128574, 142587, 142857, 258714, 258741, 285714, 285741, 412587, 412857, 425871, 428571, 1025874, 1028574, 1042587, 1042857, 1052874, 1054287, 1072854, 1074285, 1078524, 1078542, 1085274, 1085427, 1087254, 1087425, 1087524, 1087542

Offset: 1

Author

David W. Wilson

Keywords

• nonn
• base

Comments

All terms are divisible by 9. - Eric M. Schmidt, Jul 12 2014

If x and y are in the sequence, then so is 10^m*x + y if y < 10^m. - Robert Israel, Mar 20 2017

From Petros Hadjicostas, Jul 29 2020: (Start)

This is Schuh's (1968) "doubles puzzle" (the double of k is 2*k). On five pages of his book, he finds the twelve 6-digit numbers that belong to this sequence (a(2) to a(13)) and the 288 7-digit numbers of the sequence (a(14) to a(301)).

All numbers in this sequence are permutations of numbers that are combinations of numbers from A336670, which is related to another puzzle of Schuh (1968). Before he solved this puzzle, he had to solve the puzzle described in A336670.

For example, a(2) = 125874 through a(13) = 428571 are all permutations of the number 512874, which is a combination of the numbers 512 and 874 that appear in A336670. (End)

References

• Fred Schuh, The Master Book of Mathematical Recreations, Dover, New York, 1968, pp. 31-35.

Links

• David W. Wilson, Table of n, a(n) for n = 1..10001
• Mark Dominus, When do n and 2n have the same digits?

Crossrefs

Cf. A023087, A023088, A023089, A023090, A023091, A023092, A023093, A336670.

Programs

Maple

Res:= 0:
for d from 1 to 7 do
  for n from 10^(d-1)+8 to 5*10^(d-1)-1 by 9 do
    if sort(convert(n,base,10)) = sort(convert(2*n,base,10)) then
      Res:= Res, n
    fi
od od:
Res; # Robert Israel, Mar 20 2017

Mathematica

si[n_] := Sort@ IntegerDigits@ n; Flatten@ {0, Table[ Select[ Range[ 10^e+8, 5*10^e-1, 9], si[#] == si[2 #] &], {e, 6}]} (* Giovanni Resta, Mar 20 2017 *)

Python

def ok(n): return sorted(str(n)) == sorted(str(2*n))
print(list(filter(ok, range(1087543)))) # Michael S. Branicky, May 21 2021

Python

# use with ok above for larger terms
def auptod(maxd):
  return [0] + list(filter(ok, (n for d in range(2, maxd+1) for n in range(10**(d-1)-1, 5*10**(d-1), 9))))
print(auptod(7)) # Michael S. Branicky, May 22 2021