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

1,1

Links

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

Example

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

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 *)

Prog

(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

Crossrefs

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

Sequence in context: A288619 A039476 A338836 * A141965 A141975 A142663

Adjacent sequences: A061653 A061654 A061655 * A061657 A061658 A061659

Keyword

base,easy,nonn

Author

Erich Friedman, Jun 16 2001

Extensions

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

Status

approved