A031443 Digitally balanced numbers: positive numbers that in base 2 have the same number of 0's as 1's.
2, 9, 10, 12, 35, 37, 38, 41, 42, 44, 49, 50, 52, 56, 135, 139, 141, 142, 147, 149, 150, 153, 154, 156, 163, 165, 166, 169, 170, 172, 177, 178, 180, 184, 195, 197, 198, 201, 202, 204, 209, 210, 212, 216, 225, 226, 228, 232, 240, 527, 535, 539, 541, 542, 551
Offset: 1
Examples
9 is a term because '1001' contains 2 '0's and 2 '1's.
Links
- Reikku Kulon, Table of n, a(n) for n = 1..10000
- Jason Bell, Thomas F. Lidbetter and Jeffrey Shallit, Additive number theory via approximation by regular languages, International Journal of Foundations of Computer Science, Vol. 31, No. 6 (2020), pp. 667-687; arXiv preprint, arXiv:1804.07996 [cs.FL], 2018.
- Thomas Finn Lidbetter, Counting, Adding, and Regular Languages, Master's Thesis, University of Waterloo, Ontario, Canada, 2018.
- Reinhard Zumkeller, Haskell Programs for Binary Digitally Balanced Numbers.
- Index entries for sequences related to binary expansion of n
Crossrefs
Programs
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Haskell
-- See link, showing that Ulrich Schimkes formula provides a very efficient algorithm. Reinhard Zumkeller, Jun 15 2011
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Magma
[ n: n in [2..250] | Multiplicity({* z: z in Intseq(n,2) *}, 0) eq &+Intseq(n,2) ]; // Bruno Berselli, Jun 07 2011 -
Maple
a:=proc(n) local nn, n1, n0: nn:=convert(n,base,2): n1:=add(nn[i],i=1..nops(nn)): n0:=nops(nn)-n1: if n0=n1 then n else end if end proc: seq(a(n), n = 1..240); # Emeric Deutsch, Jul 31 2008
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Mathematica
Select[Range[250],DigitCount[#,2,1]==DigitCount[#,2,0]&] (* Harvey P. Dale, Jul 22 2013 *) FromDigits[#,2]&/@DeleteCases[Flatten[Permutations/@Table[PadRight[{},2n,{1,0}],{n,5}],1],?(#[[1]]==0&)]//Sort (* _Harvey P. Dale, May 30 2016 *)
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PARI
for(n=1,100,b=binary(n); l=length(b); if(sum(i=1,l, component(b,i))==l/2,print1(n,",")))
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PARI
is(n)=hammingweight(n)==hammingweight(bitneg(n,#binary(n))) \\ Charles R Greathouse IV, Mar 29 2013
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PARI
is(n)=2*hammingweight(n)==exponent(n)+1 \\ Charles R Greathouse IV, Apr 18 2020
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Perl
for my $half ( 1 .. 4 ) { my $N = 2 * $half; # only even widths apply my $vector = (1 << ($N-1)) | ((1 << ($N/2-1)) - 1); # first key my $n = 1; $n *= $_ for 2 .. $N; # N! my $d = 1; $d *= $_ for 2 .. $N/2; # (N/2)! for (1 .. $n/($d*$d*2)) { print "$vector, "; my ($v, $d) = ($vector, 0); until ($v & 1 or !$v) { $d = ($d << 1)|1; $v >>= 1 } $vector += $d + 1 + (($v ^ ($v + 1)) >> 2); # next key } } # Ruud H.G. van Tol, Mar 30 2014 -
Python
from sympy.utilities.iterables import multiset_permutations A031443_list = [int('1'+''.join(p),2) for n in range(1,10) for p in multiset_permutations('0'*n+'1'*(n-1))] # Chai Wah Wu, Nov 15 2019
Formula
a(n+1) = a(n) + 2^k + 2^(m-1) - 1 + floor((2^(k+m) - 2^k)/a(n))*(2^(2*m) + 2^(m-1)) where k is the largest integer such that 2^k divides a(n) and m is the largest integer such that 2^m divides a(n)/2^k+1. - Ulrich Schimke (UlrSchimke(AT)aol.com)
A145037(a(n)) = 0. - Reikku Kulon, Oct 02 2008
Comments