A061656 Numbers k > 1 such that, in base 2, k and k^2 contain the same digits in the same proportion.

53, 106, 211, 212, 397, 403, 417, 419, 422, 424, 437, 441, 459, 781, 794, 801, 806, 817, 833, 834, 838, 839, 841, 844, 848, 865, 874, 882, 885, 918, 979, 1481, 1549, 1562, 1565, 1571, 1573, 1585, 1588, 1589, 1602, 1612, 1613, 1634, 1637, 1665, 1666, 1667, 1668

Offset: 1

Author

Erich Friedman, Jun 16 2001

Keywords

• base
• easy
• nonn

Examples

			53 = 110101_2 and 53^2 = 101011111001_2.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A061657, A061658, A061659, A061660, A061661, A061662, A061663, A114258, A061664.

Programs

Maple

p:= n-> add(x^i, i=convert(n, base, 2)):
a:= proc(n) option remember; local k;
      for k from 1+`if`(n=1, 0, a(n-1))
      while p(k)*2<>p(k^2) do od; k
    end:
seq(a(n), n=1..50);  # Alois P. Heinz, May 10 2015

Mathematica

b2pQ[n_]:=Module[{bn=IntegerDigits[n,2],b2n=IntegerDigits[n^2,2], cbn0, cb2n0}, cbn0=Count[bn,0];cb2n0=Count[b2n,0];cbn0>0&&cb2n0>0 && Count[ bn,1]/cbn0==Count[b2n,1]/cb2n0]; Select[Range[1700],b2pQ] (* Harvey P. Dale, Jan 25 2012 *)

Python

from fractions import Fraction
from itertools import count, islice
def f(i, j):
    bi, bj = bin(i)[2:], bin(j)[2:]
    pi = [Fraction(bi.count(d), len(bi)) for d in "01"]
    pj = [Fraction(bj.count(d), len(bj)) for d in "01"]
    return pi == pj
def ok(n): return f(n, n**2)
print([k for k in range(2, 1700) if ok(k)]) # Michael S. Branicky, Feb 27 2023

Extensions

Offset changed to 1 by Alois P. Heinz, May 10 2015