A031443 -id:A031443 - cp's OEIS frontend

A071925 Digitally balanced numbers: binary numbers which have the same number of 0's as 1's; decimal representation: A031443.

10, 1001, 1010, 1100, 100011, 100101, 100110, 101001, 101010, 101100, 110001, 110010, 110100, 111000, 10000111, 10001011, 10001101, 10001110, 10010011, 10010101, 10010110, 10011001, 10011010, 10011100, 10100011

Offset: 1

Reinhard Zumkeller, Jun 14 2002

Harvey P. Dale, Table of n, a(n) for n = 1..2000

Cf. A007088, A031443.

Select[FromDigits/@Tuples[{0,1},8],DigitCount[#,10,1]==DigitCount[#,10,0]&] (* Harvey P. Dale, Sep 08 2024 *)

from itertools import islice, count
from sympy.utilities.iterables import multiset_permutations
def A071925gen(): # generator of terms
    for n in count(1):
        yield from (int('1'+''.join(p)) for p in multiset_permutations('0'*n+'1'*(n-1)))
A071925_list = list(islice(A071925gen(),30)) # Chai Wah Wu, Dec 06 2021

a(n) = A007088(A031443(n)).

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.

A345397 a(n) is the least number k such that {k, k^2, ..., k^n} are all digitally balanced numbers in base 2 (A031443), or 0 if no such k exists.

Original entry on oeis.org

2, 212, 3274, 15113, 236417, 15975465, 991017155, 4006725713, 4079348720699

Offset: 1

Amiram Eldar, Jun 17 2021

Conjecture: all a(n) > 0. - Alex Ratushnyak, Apr 26 2022

			a(1) = 2 since 2 is digitally balanced: its binary representation, 10, has the same number of 0's and 1's.
a(2) = 212 since both 212 and 212^2 are digitally balanced: the binary representation of 212, 11010100, has 4 0's and 4 1's, and the binary representation of 212^2, 1010111110010000, has 8 0's and 8 1's.

Cf. A031443, A143146, A345396.

balQ[n_] := Module[{d = IntegerDigits[n, 2], m}, EvenQ @ (m = Length @ d) && Count[d, 1] == m/2]; f[k_] := Module[{e = 0, r = k}, While[balQ[r], r *= k; e++]; e]; mx = 5; s = Table[0, {mx}]; c = 0; n = 1; While[c < mx, k = f[n]; Do[If[s[[i]] == 0, s[[i]] = n; c++], {i, 1, k}]; n++]; s

from itertools import count, islice
from sympy.utilities.iterables import multiset_permutations
def isbalanced(n): b = bin(n)[2:]; return b.count("0") == b.count("1")
def A031443gen(): yield from (int("1"+"".join(p), 2) for n in count(1) for p in multiset_permutations("0"*n+"1"*(n-1)))
def a(n):
    for k in A031443gen():
        if all(isbalanced(k**i) for i in range(2, n+1)):
            return k
print([a(n) for n in range(1, 6)]) # Michael S. Branicky, Apr 26 2022

a(9) from Bert Dobbelaere, Jun 18 2021

Name edited by Alex Ratushnyak, Apr 26 2022

A145057 First differences of A031443.

Original entry on oeis.org

2, 7, 1, 2, 23, 2, 1, 3, 1, 2, 5, 1, 2, 4, 79, 4, 2, 1, 5, 2, 1, 3, 1, 2, 7, 2, 1, 3, 1, 2, 5, 1, 2, 4, 11, 2, 1, 3, 1, 2, 5, 1, 2, 4, 9, 1, 2, 4, 8, 287, 8, 4, 2, 1, 9, 4, 2, 1, 5, 2, 1, 3, 1, 2, 11, 4, 2, 1, 5, 2, 1, 3, 1, 2, 7, 2, 1, 3, 1, 2, 5, 1, 2, 4, 15, 4, 2, 1, 5, 2, 1, 3, 1, 2, 7, 2, 1, 3, 1, 2, 5, 1

Offset: 1

Reikku Kulon, Sep 30 2008

a(n) equals the differences between k where A145037(k) equals zero. - Reikku Kulon, Oct 02 2008

			a(3) = A031443(3) - A031443(3-1) = 10 - 9 = 1. - _David A. Corneth_, Apr 26 2022

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A031443, A145037.

Differences[Select[Range[0, 1000], SameQ@@DigitCount[#, 2] &]] (* Paolo Xausa, Nov 16 2024 *)

A353139 Digitally balanced numbers (A031443) whose squares are also digitally balanced.

Original entry on oeis.org

212, 781, 794, 806, 817, 838, 841, 844, 865, 2962, 3101, 3130, 3171, 3178, 3185, 3213, 3219, 3226, 3269, 3274, 3335, 3353, 3354, 3356, 3370, 3378, 3490, 3496, 3521, 3528, 3595, 3597, 3606, 3610, 3626, 3651, 3672, 3718, 3777, 11797, 11798, 11850, 11938, 12049

Offset: 1

Alex Ratushnyak, Apr 26 2022

Numbers x such that both x and x^2 are terms of A031443, that is, have the same number of 0's as 1's in their binary representations.

Cf. A031443, A345397.

balQ[n_] := Module[{d = IntegerDigits[n, 2], m}, EvenQ @ (m = Length @ d) && Count[d, 1] == m/2]; Select[Range[12000], balQ[#] && balQ[#^2] &] (* Amiram Eldar, Apr 26 2022 *)

from itertools import count, islice
from sympy.utilities.iterables import multiset_permutations
def isbalanced(n): b = bin(n)[2:]; return b.count("0") == b.count("1")
def A031443gen(): yield from (int("1"+"".join(p), 2) for n in count(1) for p in multiset_permutations("0"*n+"1"*(n-1)))
def agen():
    for k in A031443gen():
        if isbalanced(k**2):
            yield k
print(list(islice(agen(), 40))) # Michael S. Branicky, Apr 26 2022

A337258 Primes p such that p and the prime next to p are both digitally balanced numbers in base 2 (A031443).

Original entry on oeis.org

37, 139, 557, 563, 613, 647, 653, 659, 2389, 2467, 2699, 2851, 8311, 8423, 8627, 8677, 8681, 8807, 8819, 9011, 9043, 9049, 9157, 9319, 9323, 9419, 9613, 9803, 9811, 9817, 9829, 9923, 10331, 10343, 10453, 10597, 11279, 11317, 11353, 11399, 11587, 11783, 11789

Offset: 1

Amiram Eldar, Nov 21 2020

Prime p such that p and the prime next to p are both terms of A066196.

			37 is a term since it is a prime number, and both 37 and the next prime, 41, are digitally balanced in base 2: the binary representation of 37 is 100101, the binary representation of 41 is 101001, and both contain 3 0's and 3 1's.

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

Cf. A031443, A066196, A095018, A337238.

digBalQ[n_] := Module[{d = IntegerDigits[n, 2], m}, EvenQ@(m = Length@d) && Count[d, 1] == m/2]; p = Select[Range[3*10^4], PrimeQ]; p[[Position[Partition[digBalQ /@ p, 2, 1], {True, True}] // Flatten]]

A345395 Composite numbers whose divisors that are larger than 1 are all digitally balanced numbers in base 2 (A031443).

Original entry on oeis.org

132061, 138421, 151427, 532393, 545269, 546407, 557983, 559609, 568801, 570709, 573193, 579013, 590687, 595853, 599707, 604873, 610777, 624553, 630293, 635213, 2102767, 2105063, 2109383, 2111339, 2123677, 2128187, 2129081, 2129609, 2143961, 2149753, 2151131, 2151661

Offset: 1

Amiram Eldar, Jun 17 2021

The prime numbers with this property are the digitally balanced primes (A066196).
All the terms are odd, since if k is an even digitally balanced number then its divisor k/2 is not digitally balanced (since it has one fewer 0 in its binary expansion).
Apparently most of the terms are semiprimes (A001358) with 4 divisors.
Terms with 3 divisors, i.e., squares of primes: 145178401 = 12049^2, 155575729 = 12473^2, ...
The least term with more than 4 divisors is 8897396239 = 163 * 929 * 58757, with 8 divisors.
The least term with 6 divisors is 8923691369 = 41 * 14753^2.

			132061 is a term since its divisors that are larger than 1 are {41, 3221, 132061}, and their binary representations are {101001, 110010010101, 100000001111011101}. Each one has an equal number of 0's and 1's.

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

Subsequence of A031443.

Cf. A066196.

balQ[n_] := Module[{d = IntegerDigits[n, 2], m}, EvenQ @ (m = Length @ d) && Count[d, 1] == m/2]; Select[Range[9,10^6,2], CompositeQ[#] && AllTrue[Rest@Divisors[#], balQ] &]

isbal(k) = exponent(k) + 1 == 2 * hammingweight(k);
isok(k) = if(k == 1 || isprime(k), 0, fordiv(k, d, if(d > 1 && !isbal(d), return(0))); 1); \\ Amiram Eldar, Jul 03 2025

A351599 a(n) is the smallest integer m > 0 such that m*n is a digitally balanced number (A031443).

Original entry on oeis.org

2, 1, 3, 3, 2, 2, 5, 7, 1, 1, 4, 1, 4, 3, 9, 15, 9, 10, 2, 9, 2, 2, 8, 9, 2, 2, 5, 2, 8, 5, 17, 31, 5, 5, 1, 5, 1, 1, 4, 6, 1, 1, 4, 1, 3, 4, 3, 5, 1, 1, 3, 1, 4, 4, 3, 1, 4, 4, 3, 3, 13, 9, 33, 63, 3, 3, 3, 3, 10, 3, 2, 3, 11, 9, 2, 3, 2, 2, 8, 3, 10, 9, 2

Offset: 1

Alex Ratushnyak, May 02 2022

Cf. A031443, A143146.

balQ[n_] := Module[{d = IntegerDigits[n, 2], m}, EvenQ@(m = Length@d) && Count[d, 1] == m/2]; a[n_] := Module[{k = 1}, While[!balQ[k*n], k++]; k]; Array[a, 100] (* Amiram Eldar, May 02 2022 *)

is(n) = hammingweight(n)==hammingweight(bitneg(n, #binary(n))); \\ A031443
a(n) = my(m=1); while (!is(m*n), m++); m; \\ Michel Marcus, May 02 2022

a(n) = A143146(n) / n. - Rémy Sigrist, Jul 11 2022

A357035 a(n) is the smallest number that has exactly n divisors that are digitally balanced numbers (A031443).

Original entry on oeis.org

1, 2, 10, 36, 150, 180, 420, 840, 900, 3420, 2520, 5040, 6300, 7560, 12600, 15120, 18900, 42840, 32760, 37800, 95760, 105840, 69300, 124740, 163800, 138600, 166320, 327600, 249480, 207900, 491400, 622440, 498960, 706860, 415800, 963900, 1496880, 1164240, 1081080

Offset: 0

Marius A. Burtea, Sep 20 2022

			1 has no divisors in A031443, so a(0) = 1;
2 has divisors 1 = 1_2, 2 = 10_2 and 2 = A031443(1), so a(1) = 2.
10 has divisors 2 = 10_2 and 10 = 1010_2 in A031443, so a(2) = 10.

Robert Israel, Table of n, a(n) for n = 0..97

Cf. A031443.

bal:=func; a:=[]; for n in [0..38] do k:=1; while #[d:d in Divisors(k)|bal(d)] ne n  do k:=k+1; end while; Append(~a,k); end for; a;

N:= 20: # for terms before the first term >= 2^(N+1)
W:= Vector(2^(N+1),datatype=integer[4]):
for d from 2 to N by 2 do
for t from 2^(d-1) to 2^d-1 do
  if convert(convert(t,base,2),`+`) = d/2 then
    J:= [seq(i,i=t..2^(N+1), t)];
    W[J]:= W[J] +~ 1;
fi od od:
M:= max(W);
V:= Array(0..M); count:= 0:
for i from 1 to 2^(N+1) do
  if V[W[i]] = 0 then V[W[i]]:= i; count:= count+1 fi
od:
L:= convert(V,list):
if not member(0,L,'m') then m:= M+2 fi:
L[1..m-1]; # Robert Israel, Sep 27 2023

digBalQ[n_] := Module[{d = IntegerDigits[n, 2], m}, EvenQ @ (m = Length[d]) && Count[d, 1] == m/2]; f[n_] := DivisorSum[n, 1 &, digBalQ[#] &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n] + 1; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[40, 10^7] (* Amiram Eldar, Sep 26 2022 *)

Corrected by Robert Israel, Sep 27 2023

A358857 Least integer k in A031443 such that k*n is also in A031443, or -1 if there is no such k.

Original entry on oeis.org

2, -1, 49, -1, 2, 2, 535, -1, 3843, 899, 49, 197, 12, 52, 9, -1, 9, 10, 2, 9, 2, 2, 35, 9, 2, 2, 147, 2, 2141, 2095, 32991, -1, 258055, 63495, 3849, 15367, 961, 906, 226, 3603, 56, 201, 49, 197, 49, 49, 50, 789, 56, 56, 42, 209, 50, 42, 44, 41, 10, 12, 10, 41

Offset: 1

Jeffrey Shallit, Dec 03 2022

It's clear that a(2^i)=-1 if i>0, because 2^i*n always has more 0's than n does. I do not know if every number other than a power

of 2 has such a k. Local maxima seem to be near odd powers of 2. It is not hard to show that a(2^n +- 1)!=-1 for n>=1.

			For n = 3 both 49 = [110001] and 49*3 = [10010011] have the same number of 0's as 1's, and this is the least such.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program

Cf. A031443, A358858.

PARI
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See Links section.
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from itertools import count
from sympy.utilities.iterables import multiset_permutations
def isbalanced(n): b = bin(n)[2:]; return b.count("0") == b.count("1")
def A031443gen(): yield from (int("1"+"".join(p), 2) for n in count(1) for p in multiset_permutations("0"*n+"1"*(n-1)))
def a(n):
    if n > 1 and bin(n)[2:].strip("0") == "1": return -1
    return next(k for k in A031443gen() if isbalanced(k*n))
print([a(n) for n in range(1, 61)]) # Michael S. Branicky, Dec 03 2022

a(n) = A358858(n)/n unless a(n) = -1. - Pontus von Brömssen, Dec 03 2022

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A071925 Digitally balanced numbers: binary numbers which have the same number of 0's as 1's; decimal representation: A031443.

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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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A345397 a(n) is the least number k such that {k, k^2, ..., k^n} are all digitally balanced numbers in base 2 (A031443), or 0 if no such k exists.

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A145057 First differences of A031443.

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A353139 Digitally balanced numbers (A031443) whose squares are also digitally balanced.

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A337258 Primes p such that p and the prime next to p are both digitally balanced numbers in base 2 (A031443).

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A345395 Composite numbers whose divisors that are larger than 1 are all digitally balanced numbers in base 2 (A031443).

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PARI

A351599 a(n) is the smallest integer m > 0 such that m*n is a digitally balanced number (A031443).

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