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 A061656 Numbers k > 1 such that, in base 2, k and k^2 contain the same digits in the same proportion. 12
 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified March 31 15:00 EDT 2023. Contains 361668 sequences. (Running on oeis4.)