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A061656
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Numbers k > 1 such that, in base 2, k and k^2 contain the same digits in the same proportion.
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12
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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)
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OFFSET
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1,1
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LINKS
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Alois P. Heinz, Table of n, a(n) for n = 1..10000
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EXAMPLE
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53 = 110101_2 and 53^2 = 101011111001_2.
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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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
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KEYWORD
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base,easy,nonn
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AUTHOR
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Erich Friedman, Jun 16 2001
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EXTENSIONS
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Offset changed to 1 by Alois P. Heinz, May 10 2015
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STATUS
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approved
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