A037861 -id:A037861 - cp's OEIS frontend

A385643 Galileo sequence with ratio k = 5: a(1) = 1, a(2) = k, a(2n-1) = floor(((k + 1)a(n) -1)/2), and a(2n) = floor((k + 1)a(n)/2) + 1 for n > 2.

1, 5, 14, 16, 41, 43, 47, 49, 122, 124, 128, 130, 140, 142, 146, 148, 365, 367, 371, 373, 383, 385, 389, 391, 419, 421, 425, 427, 437, 439, 443, 445, 1094, 1096, 1100, 1102, 1112, 1114, 1118, 1120, 1148, 1150, 1154, 1156, 1166, 1168, 1172, 1174, 1256, 1258, 1262

Offset: 1

Stefano Spezia, Jul 06 2025

Solution to Exercise 1.2.3 on page 35 in Tattersall.

A Galileo sequence of ratio k > 0 has the property that 1/k = a(1)/a(2) = (a(1) + a(2))/(a(3) + a(4)) = (a(1) + a(2) + a(3))/(a(4) + a(5) + a(6)) = ...

			1/5 = (1 + 5)/(14 + 16) = (1 + 5 + 14)/(16 + 41 + 43) = ...

James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 23, 35.

Stefano Spezia, Table of n, a(n) for n = 1..10000

Similar sequences for k=1..5: A037861, A385610, A005408 [Galileo, 1615], A385587, this sequence.

k=5; a[1]=1; a[2]=k; a[n_]:=a[n]=If[OddQ[n], Floor[((k+1)*a[(n+1)/2]-1)/2], Floor[(k+1)*a[n/2]/2]+1]; Array[a, 51]

A114116 1's-counting matrix: row sums give number of 1's in binary expansion of n+1.

Original entry on oeis.org

1, 0, 1, 2, -1, 1, -1, 2, -1, 1, 1, 0, 1, -1, 1, 1, 0, 0, 1, -1, 1, 3, -2, 2, -1, 1, -1, 1, -2, 3, -2, 2, -1, 1, -1, 1, 0, 1, 0, 0, 1, -1, 1, -1, 1, 0, 1, 0, 0, 0, 1, -1, 1, -1, 1, 2, -1, 2, -2, 2, -1, 1, -1, 1, -1, 1, 0, 1, -1, 2, -2, 2, -1, 1, -1, 1, -1, 1, 2, -1, 1, 0, 0, 0, 1, -1, 1, -1, 1, -1, 1, 2, -1, 1, 0, 0, 0, 0, 1, -1, 1, -1, 1, -1, 1, 4, -3, 3, -2, 2

Offset: 0

Paul Barry, Nov 13 2005

First column is -A037861(n+1). Row sums are A000120. Product of partial sum matrix (1/(1-x),x) and A114115. Inverse is A114117.

			Triangle begins
1;
0, 1;
2,-1, 1;
-1, 2,-1, 1;
1, 0, 1,-1, 1;
1, 0, 0, 1,-1, 1;
3,-2, 2,-1, 1,-1, 1;

A126388 Denominators in a series for the "alternating Euler constant" log(4/Pi).

Original entry on oeis.org

2, 3, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 22, 23, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 72, 73, 78, 79, 80, 81, 86, 87, 90, 91, 92

Offset: 2

Jonathan Sondow, Jan 01 2007

All n > 1 such that (# of 1's) != (# of 0's) in the base 2 expansion of floor(n/2). The numerators of the series are A126389.

			floor(13/2) = 6 = 110 base 2, which has (# of 1's) = 2 != 1 = (#
of 0's), so 13 is a member.

Jonathan Sondow, Double integrals for Euler's constant and ln(4/Pi) and an analog of Hadjicostas's formula, arXiv:math/0211148 [math.CA], 2002-2004.
Jonathan Sondow, Double integrals for Euler's constant and ln(4/Pi) and an analog of Hadjicostas's formula, Amer. Math. Monthly 112 (2005), 61-65.
Jonathan Sondow, New Vacca-Type Rational Series for Euler's Constant and Its "Alternating" Analog ln(4/Pi), arXiv:math/0508042 [math.NT], 2005.
Jonathan Sondow, New Vacca-Type Rational Series for Euler's Constant and Its "Alternating" Analog ln(4/Pi), Additive Number Theory, Festschrift In Honor of the Sixtieth Birthday of Melvyn B. Nathanson (D. Chudnovsky and G. Chudnovsky, eds.), Springer, 2010, pp. 331-340.
Eric Weisstein's MathWorld, Digit Count.

Complementary to A066879.

Cf. A037861, A094640, A094641, A110625, A110626, A126389.

b[n_] := DigitCount[n,2,1] - DigitCount[n,2,0]; L = {}; Do[If[b[Floor[n/2]] != 0, L = Append[L,n]], {n,2,100}]; L

log(4/Pi) = 1/2 - 1/3 + 2/6 - 2/7 - 1/8 + 1/9 + 1/10 - 1/11 + 1/12 - 1/13 + 3/14 - 3/15 - 2/16 + 2/17 + 2/22 - ...

A255568 Numbers in whose binary representation there are six 1-bits more than there are nonleading 0-bits.

Original entry on oeis.org

63, 191, 223, 239, 247, 251, 253, 254, 639, 703, 735, 751, 759, 763, 765, 766, 831, 863, 879, 887, 891, 893, 894, 927, 943, 951, 955, 957, 958, 975, 983, 987, 989, 990, 999, 1003, 1005, 1006, 1011, 1013, 1014, 1017, 1018, 1020, 2303, 2431, 2495, 2527, 2543, 2551, 2555

Offset: 1

Antti Karttunen, Mar 11 2015

Numbers for which A037861(n) = -6.

Numbers in whose binary representation (A007088) the number of 1-bits = 6 + number of (nonleading) 0 bits.

			63 ("111111" in binary) is included because there are 0 zero-bits and six 1-bits.
191 ("10111111" in binary) is included because there is 1 zero-bit and seven 1-bits, thus there are six 1-bits more than the number of 0-bits.

Antti Karttunen, Table of n, a(n) for n = 1..16303

Cf. A007088, A037861.

The intersection of A030130 and A023689 is a finite subsequence of this sequence.

is(n)=2*hammingweight(n)==6+#binary(n) \\ Charles R Greathouse IV, Dec 16 2015

for(b=3,6, for(n=2^(2*b-1),4^b-1, if(hammingweight(n)==3+b, print1(n", ")))) \\ Charles R Greathouse IV, Dec 16 2015

listBBitMembers(b)=if(b%2,return([])); my(u=List()); forvec(v=vector(3+b/2,i,[0,b-1]), listput(u, sum(i=1,#v,2^v[i])),2); Vec(u) \\ Charles R Greathouse IV, Dec 16 2015

use ntheory ":all"; my $bits = 0; for (1..1000) { my $o = hammingweight($); $bits++ if ($ & ($-1))==0; say if $o == 6+$bits-$o; } # _Dana Jacobsen, Dec 16 2015

A258001 Those terms of A255571 whose every A080541/A080542-rotation is also a term of A255571.

Original entry on oeis.org

1, 64, 127, 1057, 1090, 1156, 1288, 1519, 1552, 1783, 1915, 1981, 2014, 4369, 4642, 5188, 6007, 6280, 7099, 7645, 7918, 16963, 17029, 17161, 17542, 17545, 17674, 17938, 18529, 18577, 18700, 18706, 18964, 19492, 20335, 20641, 20674, 20770, 21016, 21028, 21544, 22447, 22600, 23479, 23503, 23995, 24187, 24253, 24286, 24865, 24898, 24964

Offset: 0

Antti Karttunen, May 16 2015

These are the numbers whose binary representation traces a nonselfcrossing circuit in honeycomb lattice when its bits (from the least to the second most significant bit; the most significant 1-bit is ignored) are interpreted as directions to proceed at each vertex, with an additional condition that the final direction (angle) must be equal to the starting direction of the walk. Because each bit either adds or subtracts 60 degrees from the current phase angle, it implies that for all terms after the initial term a(1)=0 (which stands for an empty path), the difference between the number of 0-bits and 1-bits (when excluding the most significant bit which is always 1) must be either -6 or +6. And indeed, for all n >= 1, A037861(a(n)) is either 5 or -7 as A037861 takes also the most significant bit into account.

Antti Karttunen, Table of n, a(n) for n = 0..13242 (all terms up to binary width 24+1)

Subsequence of A255571.
Cf. A037861, A080541, A080542.
Cf. A258002 (a subsequence; terms that have more ones than zeros in their binary representation).

A031461 Numbers whose base-4 representation has the same number of 0's as 3's.

Original entry on oeis.org

1, 2, 5, 6, 9, 10, 12, 19, 21, 22, 25, 26, 28, 35, 37, 38, 41, 42, 44, 49, 50, 52, 56, 71, 75, 77, 78, 83, 85, 86, 89, 90, 92, 99, 101, 102, 105, 106, 108, 113, 114, 116, 120, 135, 139, 141, 142, 147, 149, 150, 153, 154, 156, 163, 165

Offset: 1

Clark Kimberling

Union of A031443 and A031448, i.e., numbers k in binary where the difference between numbers of 1's and 0's [A000120(k) - A023416(k) = -A037861(k)] is 1 or 0. - Henry Bottomley, Mar 02 2001

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

Cf. A000120, A007090, A023416, A031443, A031448, A037861.

Select[Range[165], Equal @@ DigitCount[#, 4, {0, 3}] &] (* Amiram Eldar, Aug 03 2023 *)

A126389 Numerators in a series for the "alternating Euler constant" log(4/Pi).

Original entry on oeis.org

1, -1, 2, -2, -1, 1, 1, -1, 1, -1, 3, -3, -2, 2, 2, -2, 2, -2, 2, -2, 4, -4, -3, 3, -1, 1, -1, 1, 1, -1, -1, 1, 1, -1, 1, -1, 3, -3, -1, 1, 1, -1, 1, -1, 3, -3, 1, -1, 3, -3, 3, -3, 5, -5, -4, 4, -2, 2, -2, 2, -2, 2, 2, -2, -2, 2, 2, -2, 2, -2, 2, -2, 4, -4, -2, 2

Offset: 2

Jonathan Sondow, Jan 01 2007

Nonzero values of (-1)^n*b(floor(n/2)) for n > 1, where b(n) = (# of 1's) - (# of 0's) in the base 2 expansion of n. The denominators of the series are A126388.

			floor(15/2) = 7 = 111 base 2, which has (# of 1's) - (# of 0's) = 3, so (-1)^15*3 = -3 is a term.

J. Sondow, Double integrals for Euler's constant and ln(4/Pi) and an analog of Hadjicostas's formula, Amer. Math. Monthly 112 (2005) 61-65.
J. Sondow, New Vacca-Type Rational Series for Euler's Constant and Its "Alternating" Analog ln(4/Pi), Additive Number Theory, Festschrift In Honor of the Sixtieth Birthday of Melvyn B. Nathanson (D. Chudnovsky and G. Chudnovsky, eds.), Springer, 2010, pp. 331-340.
Eric Weisstein's MathWorld, Digit Count

Cf. A037861, A066879, A094640, A094641, A110625, A110626, A126388.

b[n_] := DigitCount[n,2,1] - DigitCount[n,2,0]; L = {}; Do[If[b[Floor[n/2]] != 0, L = Append[L,(-1)^n*b[Floor[n/2]]]], {n,2,100}]; L

Log(4/Pi) = 1/2 - 1/3 + 2/6 - 2/7 - 1/8 + 1/9 + 1/10 - 1/11 + 1/12 - 1/13 + 3/14 - 3/15 - 2/16 + 2/17 + 2/22 - ...

A143903 A positive integer k is included if |(number of 0's in binary k) - (number of 1's in binary k)| divides k.

Original entry on oeis.org

1, 4, 5, 6, 8, 14, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 34, 36, 40, 46, 48, 54, 58, 60, 66, 67, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 92, 93, 95, 96, 97, 98, 99, 100, 101, 102, 104, 105, 106, 108, 112, 113, 114, 116

Offset: 1

Leroy Quet, Sep 04 2008

nonn

			27 in binary is 11011. The number of 0's is 1. The number of 1's is 4. Since |1-4|=3 divides 27, 27 is a term of the sequence.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A037861.

Select[Range[100], Abs[DigitCount[ #, 2, 1] - DigitCount[ #, 2, 0]] > 0 && Mod[ #, Abs[DigitCount[ #, 2, 1] - DigitCount[ #, 2, 0]]] == 0 &] (* Stefan Steinerberger, Sep 05 2008 *)

More terms from Stefan Steinerberger, Sep 05 2008

A303065 Number of numbers < n whose binary representation has the same difference between the numbers of 0's and 1's as n does.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 0, 0, 1, 2, 1, 3, 2, 3, 0, 0, 2, 3, 3, 4, 4, 5, 1, 5, 6, 7, 2, 8, 3, 4, 0, 0, 1, 2, 4, 3, 5, 6, 4, 4, 7, 8, 5, 9, 6, 7, 1, 5, 10, 11, 8, 12, 9, 10, 2, 13, 11, 12, 3, 13, 4, 5, 0, 0, 1, 2, 6, 3, 7, 8, 9, 4, 9, 10, 10, 11, 11, 12, 5, 5, 12, 13

Offset: 0

Alex Ratushnyak, Apr 17 2018

First occurrence of k, k=0,1,2,...: 0, 4, 6, 12, 20, 22, 25, 26, 28, 44, 49, ..., . - Robert G. Wilson v, Feb 08 2018

Ordinal transform of A037861, minus one. - David Radcliffe, May 21 2025

			There are two numbers below 6 with number of 1's in the binary representation minus number of 0's equal to 1, namely 1 and 5, therefore a(6)=2.
There are 3 numbers below 12 with number of 1's in the binary representation minus number of 0's equal to 0, namely 2, 9, 10, therefore a(12)=3.

Alois P. Heinz, Table of n, a(n) for n = 0..32768

Cf. A037861, A097110.

b:= n-> `if`(n=0, 1, add(1-2*i, i=Bits[Split](n))):
p:= proc() -1 end:
a:= proc(n) option remember; local t;
      t:= b(n); p(t):= p(t)+1
    end:
seq(a(n), n=0..82);  # Alois P. Heinz, May 21 2025

d[n_] := DigitCount[n, 2, 1] - DigitCount[n, 2, 0]; f[n_] := Block[{fd = d[n], c = k = 0}, While[k < n, If[d@ k == fd, c++]; k++]; c]; Array[f, 83, 0] (* Robert G. Wilson v, Feb 08 2018 *)

d=[0]*200
for n in range(1024):
    b = bin(n)[2:]
    c0 = b.count('0')
    c1 = len(b) - c0
    diff = c0 - c1
    print(d[100+diff], end=', ')
    d[100+diff] += 1

from collections import Counter
from itertools import count, islice
def a303065_gen():
    counter = Counter()
    for n in count():
        bitstring = format(n, 'b')
        diff = bitstring.count('1') - bitstring.count('0')
        yield counter[diff]
        counter[diff] += 1
a303065_list = list(islice(a303065_gen(), 83)) # David Radcliffe, May 21 2025

a(n) = 0 iff n belongs to A097110. - Rémy Sigrist, May 16 2018

Offset corrected by David Radcliffe, May 21 2025

A344985 Sum of imbalance of all initial subsequences of the binary representation of n.

Original entry on oeis.org

1, 1, 3, 2, 2, 4, 6, 4, 2, 2, 4, 4, 6, 8, 10, 7, 5, 3, 3, 3, 3, 5, 7, 5, 5, 7, 9, 9, 11, 13, 15, 11, 9, 7, 5, 5, 3, 3, 5, 5, 3, 3, 5, 5, 7, 9, 11, 7, 5, 5, 7, 7, 9, 11, 13, 9, 11, 13, 15, 15, 17, 19, 21, 16, 14, 12, 10, 10, 8, 6, 6, 8, 6, 4, 4, 4, 4, 6, 8, 8

Offset: 1

André Engels, Jun 04 2021

The imbalance of a sequence of zeros and ones is given as the absolute value of the difference between the number of zeros and the number of ones (that is, the absolute value of A037861 if n is the sequence A007088).

			9 in binary is 1001. The imbalance of 1 is 1, the imbalance of 10 is 0, the imbalance of 100 is 1 and the imbalance of 1001 is 0. thus a(9)=1+0+1+0=2.

Alois P. Heinz, Table of n, a(n) for n = 1..16384

Cf. A007088, A037861.

imbal := proc(L)
     abs(numboccur(0, L) - numboccur(1, L))
end proc:
A344985 := proc(n)
    dgs := ListTools[Reverse](convert(n,base,2)) ;
    add(imbal( dgs[1..l]),l=1..nops(dgs)) ;
end proc: # R. J. Mathar, Jul 07 2021
# second Maple program:
a:= proc(n) option remember; `if`(n=0, 0, a(iquo(n, 2))+
      (l-> abs(add(1-2*i, i=l)))(Bits[Split](n)))
    end:
seq(a(n), n=1..80);  # Alois P. Heinz, Jul 07 2021

a(n) = my(b=binary(n)); sum(k=1, #b, abs(sum(j=1, k, b[j]==1) - sum(j=1, k, b[j]==0))); \\ Michel Marcus, Jun 09 2021

def A344985(n):
    s, c, b = bin(n)[2:], 0, 0
    for x in s:
        b += 1 if x == '1' else -1
        c += abs(b)
    return c # Chai Wah Wu, Jul 07 2021

a(n) = a(floor(n/2)) + |A037861(n)|. - R. J. Mathar, Jul 07 2021

cp's OEIS Frontend

A385643 Galileo sequence with ratio k = 5: a(1) = 1, a(2) = k, a(2*n-1) = floor(((k + 1)*a(n) -1)/2), and a(2*n) = floor((k + 1)*a(n)/2) + 1 for n > 2.

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A114116 1's-counting matrix: row sums give number of 1's in binary expansion of n+1.

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A126388 Denominators in a series for the "alternating Euler constant" log(4/Pi).

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A255568 Numbers in whose binary representation there are six 1-bits more than there are nonleading 0-bits.

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PARI

PARI

PARI

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A258001 Those terms of A255571 whose every A080541/A080542-rotation is also a term of A255571.

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A031461 Numbers whose base-4 representation has the same number of 0's as 3's.

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Mathematica

A126389 Numerators in a series for the "alternating Euler constant" log(4/Pi).

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Keywords

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Programs

Mathematica

Formula

A143903 A positive integer k is included if |(number of 0's in binary k) - (number of 1's in binary k)| divides k.

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A303065 Number of numbers < n whose binary representation has the same difference between the numbers of 0's and 1's as n does.

A385643 Galileo sequence with ratio k = 5: a(1) = 1, a(2) = k, a(2n-1) = floor(((k + 1)a(n) -1)/2), and a(2n) = floor((k + 1)a(n)/2) + 1 for n > 2.