A191292 -id:A191292 - cp's OEIS frontend

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

Clark Kimberling

Also numbers k such that the binary digital mean dm(2, k) = (Sum_{i=1..d} 2*d_i - 1) / (2*d) = 0, where d is the number of digits in the binary representation of k and d_i the individual digits. - Reikku Kulon, Sep 21 2008
From Reikku Kulon, Sep 29 2008: (Start)
Each run of values begins with 2^(2k + 1) + 2^(k + 1) - 2^k - 1. The initial values increase according to the sequence {2^(k - 1), 2^(k - 2), 2^(k - 3), ..., 2^(k - k)}.
After this, the values follow a periodic sequence of increases by successive powers of two with single odd values interspersed.
Each run ends with an odd increase followed by increases of {2^(k - k), ..., 2^(k - 2), 2^(k - 1), 2^k}, finally reaching 2^(2k + 2) - 2^(k + 1).
Similar behavior occurs in other bases. (End)
Numbers k such that A000120(k)/A070939(k) = 1/2. - Ctibor O. Zizka, Oct 15 2008
Subsequence of A053754; A179888 is a subsequence. - Reinhard Zumkeller, Jul 31 2010
A000120(a(n)) = A023416(a(n)); A037861(a(n)) = 0.
A001700 gives number of terms having length 2*n in binary representation: A001700(n-1) = #{m: A070939(a(m))=2*n}. - Reinhard Zumkeller, Jun 08 2011
The number of terms below 2^k is A079309(floor(k/2)) for k > 1. - Amiram Eldar, Nov 21 2020

			9 is a term because '1001' contains 2 '0's and 2 '1's.

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

Subsequences: A144798, A144799, A144800, A144801, A144812, A179888.
Subsequence of A053754.
Row n = 2 of A378000.
Cf. A049354-A049360, A000120, A001316, A006519, A023416, A070939, A144777, A145057, A145058, A145059, A145060, A144912, A145037, A191292, A090050, A014486, A061854, A037861, A079309.
Terms of binary width n are enumerated by A001700.

-- See link, showing that Ulrich Schimkes formula provides a very efficient algorithm. Reinhard Zumkeller, Jun 15 2011

[ n: n in [2..250] | Multiplicity({* z: z in Intseq(n,2) *}, 0) eq &+Intseq(n,2) ];  // Bruno Berselli, Jun 07 2011

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

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 *)

for(n=1,100,b=binary(n); l=length(b); if(sum(i=1,l, component(b,i))==l/2,print1(n,",")))

is(n)=hammingweight(n)==hammingweight(bitneg(n,#binary(n))) \\ Charles R Greathouse IV, Mar 29 2013

is(n)=2*hammingweight(n)==exponent(n)+1 \\ Charles R Greathouse IV, Apr 18 2020

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

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

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

A337238 Number k such that k and k+1 are both digitally balanced numbers in base 2 (A031443).

Original entry on oeis.org

9, 37, 41, 49, 141, 149, 153, 165, 169, 177, 197, 201, 209, 225, 541, 557, 565, 569, 589, 597, 601, 613, 617, 625, 653, 661, 665, 677, 681, 689, 709, 713, 721, 737, 781, 789, 793, 805, 809, 817, 837, 841, 849, 865, 901, 905, 913, 929, 961, 2109, 2141, 2157, 2165

Offset: 1

Amiram Eldar, Nov 21 2020

All the terms are of the form 4*k + 1, where k is a digitally balanced number in base 2. Therefore, there are no 3 consecutive numbers that are digitally balanced in base 2.

The number of terms below 2^k is A079309(floor(k/2)-1) for k > 3.

			9 is a term since the binary representation of 9 is 1001, which contains 2 0's and 2 1's, and the binary representation of 9 + 1 = 10 is 1010, which also contains 2 0's and 2 1's.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A031443, A079309, A191292.

A206374 \ {2} is a subsequence.

digBalQ[n_] := Module[{d = IntegerDigits[n, 2], m}, EvenQ@(m = Length@d) && Count[d, 1] == m/2]; Select[Range[2000], digBalQ[#] && digBalQ[# + 1] &]

a(n) = 4*A031443(n) + 1.

A191341 a(n) = 4^n - 2*2^n + 3.

Original entry on oeis.org

3, 11, 51, 227, 963, 3971, 16131, 65027, 261123, 1046531, 4190211, 16769027, 67092483, 268402691, 1073676291, 4294836227, 17179607043, 68718952451, 274876858371, 1099509530627, 4398042316803, 17592177655811, 70368727400451, 281474943156227, 1125899839733763

Offset: 1

Juri-Stepan Gerasimov, May 30 2011

Also the number of dominating sets for the complete bipartite graph K_{n,n}. - Eric W. Weisstein, Apr 24 2017

For n > 1, a(n) is the largest integer such that the binary representations of a(n)-1 and a(n)+1 both contain exactly n 0's and n 1's.

Vincenzo Librandi, Table of n, a(n) for n = 1..500 (corrected by Ray Chandler, Jan 19 2019)
Eric Weisstein's World of Mathematics, Dominating Set
Eric Weisstein's World of Mathematics, Complete Bipartite Graph
Index entries for linear recurrences with constant coefficients, signature (7,-14,8).

Cf. A031443 (digitally balanced numbers), A191292, A191296.

List([1..30], n -> (2^n-1)^2+2); # G. C. Greubel, Feb 10 2019

[4^n - 2^(n+1) + 3: n in [1..30]]; // Vincenzo Librandi, Jun 09 2011

Table[3 - 2^(1+n) + 4^n, {n, 20}] (* Eric W. Weisstein, Apr 24 2017 *)
With[{r = Range[10^5]}, 2 + SplitBy[Cases[Transpose[{Partition[Tally[#][[All, 2]] & /@ IntegerDigits[r, 2], 2, 1, 1], r}], {{{n_, n_}, {n_, n_}}, p_} :> {n, p}], First][[All, -1, -1]]] (* Eric W. Weisstein, Apr 24 2017 *)
LinearRecurrence[{7, -14, 8}, {3, 11, 51}, 20] (* Eric W. Weisstein, Jun 29 2017 *)
CoefficientList[Series[(3-10x+16x^2)/((1-x)(1-2x)(1-4x)), {x, 0, 20}], x] (* Eric W. Weisstein, Jun 29 2017 *)

a(n)=4^n-2*2^n+3 \\ Charles R Greathouse IV, Jun 08 2011

[(2^n-1)^2+2 for n in (1..30)] # G. C. Greubel, Feb 10 2019

a(n) = 4^n - 2^(n+1) + 3. - Nathaniel Johnston, May 30 2011
a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3). - Eric W. Weisstein, Jun 29 2017
G.f.: (3 -10*x +16*x^2)/((1-x)*(1-2*x)*(1-4*x)). - R. J. Mathar, Jun 02 2011 ( corrected by G. C. Greubel, Feb 10 2019 )
E.g.f.: -2 + 3*exp(x) - 2*exp(2*x) + exp(4*x). - G. C. Greubel, Feb 10 2019

Definition changed to closed-form formula and original definition clarified and moved to comment by Eric W. Weisstein, Apr 24 2017

A191296 Least k such that k-1 and k+1 in binary representation have same number n of 0's as 1's.

Original entry on oeis.org

11, 36, 140, 540, 2108, 8316, 33020, 131580, 525308, 2099196, 8392700, 33562620, 134234108, 536903676, 2147549180, 8590065660, 34360000508, 137439477756, 549756862460, 2199025352700, 8796097216508, 35184380477436, 140737505132540, 562949986975740, 2251799880794108, 9007199388958716, 36028797287399420

Offset: 2

Juri-Stepan Gerasimov, May 29 2011

Colin Barker, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (7,-14,8).

Cf. A031443 (digitally balanced numbers), A191292, A191341.

Join[{11},LinearRecurrence[{7,-14,8},{36,140,540},40]] (* Harvey P. Dale, Jun 10 2011 *)

a(n)=if(n<3,11,2*(2^(n-1) + 2)*(2^(n-1) - 1)) \\ Charles R Greathouse IV, Jun 01 2011

Vec(x^2*(11 - 41*x + 42*x^2 - 24*x^3) / ((1 - x)*(1 - 2*x)*(1 - 4*x)) + O(x^40)) \\ Colin Barker, Jan 26 2018

a(n) = 2*(2^(n-1) + 2)*(2^(n-1) - 1) for n>=3. - Nathaniel Johnston, May 30 2011
a(0)=11, a(1)=36, a(2)=140, a(3)=540, a(n)=7*a(n-1)-14*a(n-2)+8*a(n-3). - Harvey P. Dale, Jun 10 2011
G.f.: x^2*(11 - 41*x + 42*x^2 - 24*x^3) / ((1 - x)*(1 - 2*x)*(1 - 4*x)). - Colin Barker, Jan 26 2018

a(11)-a(27) and recurrence from Charles R Greathouse IV, May 29 2011

A191369 Numbers n with k divisors such that n-1 and n+1 in binary representation have same number k of 0's as 1's.

Original entry on oeis.org

11, 155, 203, 2164, 2228, 2252, 2276, 2348, 2404, 2452, 2468, 2588, 2612, 2636, 2644, 2675, 2708, 2763, 2836, 2891, 2979, 3148, 3179, 3236, 3275, 3283, 3411, 3475, 3716, 3971, 33723, 33755, 34235, 34523, 34539, 34715, 34771, 35315, 35563, 35571, 35787, 36155, 36411, 36507, 36555, 36579

Offset: 1

Juri-Stepan Gerasimov, Jun 04 2011

nonn

Cf. A030513, A030516, A030516, A191292.

isA191369(n)=my(b=vecsort(binary(n-1)),k=#b\2); #b==k+k & !b[k] & b[k+1] & b==vecsort(binary(n+1)) & numdiv(n)==k \\ Charles R Greathouse IV, Jun 05 2011

a(10) corrected, a(31)-a(46) added by Charles R Greathouse IV, Jun 05 2011

A191534 Least k with 2n divisors such that k-1 and k+1 in binary representation have same number 2n of 0's as 1's.

Original entry on oeis.org

11, 155, 2164, 33723, 539379, 8396540, 136109403, 2147745531, 34360623100, 549771505659, 8797030442667, 140737513521148, 2251823188540923, 36028801313906427, 576460760876579772

Offset: 1

Juri-Stepan Gerasimov, Jun 05 2011

Does a(n) exist for every n? It seems plausible at first glance; asymptotically there should be enough numbers in the range 16^n * [1/2, 1] that have 2n divisors (since 16 > e). [Charles R Greathouse IV, Jun 05 2011]

a(16) <= 9223372071079772155. - Donovan Johnson, Sep 25 2011

Cf. A000005, A191292, A191369.

a(n)=my(v=vector(4*n,i,i>2*n));for(k=1<<(4*n-1)+1<<(2*n-1)-1,1<<(4*n)-1<<(2*n),if(vecsort(binary(k-1))==v & vecsort(binary(k+1))==v & numdiv(k)==2*n, return(k))) \\ Charles R Greathouse IV, Jun 05 2011

a(5)-a(11) from Charles R Greathouse IV, Jun 05 2011

a(12)-a(15) from Donovan Johnson, Sep 25 2011

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A031443 Digitally balanced numbers: positive numbers that in base 2 have the same number of 0's as 1's.

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A337238 Number k such that k and k+1 are both digitally balanced numbers in base 2 (A031443).

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A191341 a(n) = 4^n - 2*2^n + 3.

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A191296 Least k such that k-1 and k+1 in binary representation have same number n of 0's as 1's.

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A191369 Numbers n with k divisors such that n-1 and n+1 in binary representation have same number k of 0's as 1's.

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A191534 Least k with 2n divisors such that k-1 and k+1 in binary representation have same number 2n of 0's as 1's.

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A191556 Largest k with 2n divisors such that the binary representations of k-1 and k+1 are in base 2 each digitally balanced with 4n digits.

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