A139370 -id:A139370 - cp's OEIS frontend

A152715 Primes in A065049 which are not in A139370.

277, 337, 349, 373, 853, 1093, 1109, 1117, 1237, 1297, 1301, 1303, 1361, 1367, 1373, 1381, 1399, 1429, 1489, 1493, 1621, 1861, 1873, 1877, 1879, 2389, 3413, 3541, 4177, 4357, 4373, 4421, 4423, 4441, 4447, 4549, 4561, 4567, 4597, 4933, 4951, 4957, 5077, 5189, 5197, 5209, 5233, 5237

Offset: 1

Vladimir Shevelev, Dec 11 2008, Dec 12 2008

nonn

In the notation of A139370, a prime p is in the sequence iff e(p)>o(p) and e(p)-o(p)== 4 or 5 (mod 6). [Vladimir Shevelev, Dec 12 2008]

Cf. A065049, A139370.

aQ[n_] := PrimeQ[n] && EvenQ[Count[IntegerDigits[n, 2], 1]] == OddQ[Mod[n, 3]] && Module[{d = Reverse[IntegerDigits[n, 2]]}, Total@d[[1;; -1;; 2]] >= Total@d[[2;; -1;; 2]]]; Select[Range[5300], aQ] (* Amiram Eldar, Dec 15 2018 *)

isokp(p) = (p>2) && isprime(p) && ((hammingweight(p) % 2) != ((p % 3) % 2)); \\ A065049
isok0(n) = {my(irb = Vec(select(x->(x%2), Vecrev(binary(n)), 1))); #select(x->(x%2), irb) < #irb/2;} \\ A139370
isok(p) = isokp(p) && !isok0(p); \\ Michel Marcus, Dec 15 2018

Missing 853 and more terms from Michel Marcus, Dec 15 2018

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219

Offset: 1

Olivier Gérard

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.
Cf. A000302, A000346, A003242, A029837, A053738, A053754, A071321, A103919, A191232, A328596.

Fortran
```
c See link in A139351.
```

N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N-1):
d:= ceil(log[4](N));
S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
for i from 1 to d do
  B:= B + S;
  S:= Array(0..N-1,i-> S[floor(i/4)]);
od:
select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)

for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002

That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

A139351 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives e(n).

Original entry on oeis.org

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

Offset: 0

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n)+o(n) = A000120(n), the binary weight of n.

a(n) is also number of 1's and 3's in 4-ary representation of n. - Frank Ruskey, May 02 2009

			For n = 43 = 2^0 + 2^1 + 2^3 + 2^5, e(43)=1, o(43)=3.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Franklin T. Adams-Watters and Frank Ruskey, Generating Functions for the Digital Sum and Other Digit Counting Sequences, JIS 12 (2009), Article 09.5.6.
N. J. A. Sloane, Fortran program for this and related sequences.

Cf. A000120, A030308, A059841, A139352, A139353, A139354, A139355, A039004, A139370, A139371, A139372, A139373.

Fortran
```
See Sloane link.
```

import Data.List (unfoldr)
a139351 = sum . map (`mod` 2) .
   unfoldr (\x -> if x == 0 then Nothing else Just (x, x `div` 4))
-- Reinhard Zumkeller, Apr 22 2011

A139351 := proc(n)
    local a,bdgs,r;
    a := 0 ;
    bdgs := convert(n,base,2) ;
    for r from 1 to nops(bdgs) by 2 do
        if op(r,bdgs) = 1 then
            a := a+1 ;
        end if;
    end do:
    a;
end proc: # R. J. Mathar, Jul 21 2016

terms = 99; s = (1/(1-z))*Sum[z^(4^m)/(1+z^(4^m)), {m, 0, Log[4, terms] // Ceiling}] + O[z]^terms; CoefficientList[s, z] (* Jean-François Alcover, Jul 21 2017 *)
a[0] = 0; a[n_] := a[n] = a[Floor[n/4]] + If[OddQ[Mod[n, 4]], 1, 0]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)

a(n)=if(n>3,a(n\4))+n%2 \\ Charles R Greathouse IV, Apr 21 2016

a(n) + A139352(n) = A000120(n).
G.f.: (1/(1-z))*Sum_{m>=0} (z^(4^m)/(1+z^(4^m))). - Frank Ruskey, May 03 2009
Recurrence relation: a(0)=0, a(4m) = a(4m+2) = a(m), a(4m+1) = a(4m+3) = 1+a(m). - Frank Ruskey, May 11 2009
a(n) = Sum_{k} A030308(n,k)*A059841(k). - Philippe Deléham, Oct 14 2011

Typo in example fixed by Reinhard Zumkeller, Apr 22 2011

A139352 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives o(n).

Original entry on oeis.org

0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 2, 3, 3, 2, 2, 3, 3, 1, 1, 2, 2, 1, 1, 2, 2, 2, 2, 3, 3, 2, 2, 3, 3, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2

Offset: 0

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n) + o(n) = A000120(n), the binary weight of n.

a(n) is also the number of 2's and 3's in the 4-ary representation of n. - Frank Ruskey, May 02 2009

			For n = 43 = 2^0 + 2^1 + 2^3 + 2^5, e(43)=1, o(43)=3. [Typo fixed by _Reinhard Zumkeller_, Apr 22 2011]

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Franklin T. Adams-Watters and Frank Ruskey, Generating Functions for the Digital Sum and Other Digit Counting Sequences, JIS 12 (2009), Article 09.5.6.

Cf. A000035, A000120, A030308, A039004, A139351, A139353, A139354, A139355, A139370, A139371, A139372, A139373.

Fortran
```
c See link in A139351
```

import Data.List (unfoldr)
a139352 = sum . map ((`div` 2) . (`mod` 4)) .
   unfoldr (\x -> if x == 0 then Nothing else Just (x, x `div` 4))
-- Reinhard Zumkeller, Apr 22 2011

A139352 := proc(n)
    local a,bdgs,r;
    a := 0 ;
    bdgs := convert(n,base,2) ;
    for r from 2 to nops(bdgs) by 2 do
        if op(r,bdgs) = 1 then
            a := a+1 ;
        end if;
    end do:
    a;
end proc: # R. J. Mathar, Jul 21 2016

a[n_] := Count[Position[Reverse@IntegerDigits[n, 2], 1]-1, {_?OddQ}];
Table[a[n], {n, 0, 99}] (* Jean-François Alcover, Mar 04 2023 *)
a[0] = 0; a[n_] := a[n] = a[Floor[n/4]] + If[Mod[n, 4] > 1, 1, 0]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)

a(n)=if(n>3,a(n\4))+n%4\2 \\ Charles R Greathouse IV, Apr 21 2016

G.f.: (1/(1-z))*Sum_{m>=0} (z^(2*4^m)/(1+(2*4^m))). - Frank Ruskey, May 03 2009
Recurrence relation: a(0)=0, a(4m) = a(4m+1) = a(m), a(4m+2) = a(4m+3) = 1+a(m). - Frank Ruskey, May 11 2009
a(n) = Sum_{k} A030308(n,k)*A000035(k). - Philippe Deléham, Oct 14 2011

A139353 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives e(n)*o(n).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 2, 0, 1, 0, 2, 1, 2, 2, 4, 0, 0, 1, 2, 0, 0, 2, 3, 1, 2, 2, 4, 2, 3, 4, 6, 0, 1, 0, 2, 1, 2, 2, 4, 0, 2, 0, 3, 2, 4, 3, 6, 1, 2, 2, 4, 2, 3, 4, 6, 2, 4, 3, 6, 4, 6, 6, 9, 0, 0, 1, 2, 0, 0, 2, 3, 1, 2, 2, 4, 2, 3, 4, 6, 0, 0, 2, 3, 0, 0, 3, 4, 2, 3, 4, 6, 3, 4, 6, 8, 1, 2, 2

Offset: 0

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n) + o(n) = A000120(n), the binary weight of n.

			If n = 43 = 2^0+2^2+2^3+2^5, e(43)=1, o(43)=3.

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A000120, A139351, A139352, A139354, A139355, A039004, A139370, A139371, A139372, A139373.

Fortran
```
c See link in A139351
```

e[0] = 0; e[n_] := e[n] = e[Floor[n/4]] + If[OddQ[Mod[n, 4]], 1, 0];
o[0] = 0; o[n_] := o[n] = o[Floor[n/4]] + If[Mod[n, 4] > 1, 1, 0];
a[n_] := e[n] * o[n]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)

a(n) = A139351(n) * A139352(n). - Amiram Eldar, Jul 18 2023

A139354 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives min{e(n), o(n)}.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 2, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 0, 1, 0, 1, 1, 1, 1, 2, 0, 1, 0, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 2, 2, 3, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 1

Offset: 0

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n) + o(n) = A000120(n), the binary weight of n.

			If n = 43 = 2^0+2^2+2^3+2^5, e(43)=1, o(43)=3.

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A000120, A139351, A139352, A139353, A139355, A039004, A139370, A139371, A139372, A139373.

Fortran
```
c See link in A139351
```

e[0] = 0; e[n_] := e[n] = e[Floor[n/4]] + If[OddQ[Mod[n, 4]], 1, 0];
o[0] = 0; o[n_] := o[n] = o[Floor[n/4]] + If[Mod[n, 4] > 1, 1, 0];
a[n_] := Min[e[n], o[n]]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)

a(n) = min(A139351(n), A139352(n)). - Amiram Eldar, Jul 18 2023

A139355 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives max{e(n), o(n)}.

Original entry on oeis.org

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

Offset: 0

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n) + o(n) = A000120(n), the binary weight of n.

			If n = 43 = 2^0+2^2+2^3+2^5, e(43)=1, o(43)=3.

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A000120, A139351, A139352, A139353, A139354, A039004, A139370, A139371, A139372, A139373.

Fortran
```
c See link in A139351
```

e[0] = 0; e[n_] := e[n] = e[Floor[n/4]] + If[OddQ[Mod[n, 4]], 1, 0];
o[0] = 0; o[n_] := o[n] = o[Floor[n/4]] + If[Mod[n, 4] > 1, 1, 0];
a[n_] := Max[e[n], o[n]]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)

a(n) = max(A139351(n), A139352(n)). - Amiram Eldar, Jul 18 2023

A139371 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence lists n such that e(n) <= o(n).

Original entry on oeis.org

0, 2, 3, 6, 8, 9, 10, 11, 12, 14, 15, 18, 24, 26, 27, 30, 32, 33, 34, 35, 36, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 54, 56, 57, 58, 59, 60, 62, 63, 66, 72, 74, 75, 78, 90, 96, 98, 99, 102, 104, 105, 106, 107, 108, 110, 111, 114, 120, 122, 123, 126

Offset: 1

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n)+o(n) = A000120(n), the binary weight of n. For e(n) = o(n) see A039004.

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

Cf. A000120, A139351, A139352, A139353, A139354, A139355, A039004, A139370, A139372, A139373.

Fortran
```
c See link in A139351
```

q[n_] := Module[{d = Reverse[IntegerDigits[n, 2]]}, Total@ d[[1;; -1;; 2]] <= Total@ d[[2;; -1;; 2]]]; Select[Range[0, 130], q] (* Amiram Eldar, Aug 31 2023 *)

A139372 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence lists n such that e(n) >= o(n).

Original entry on oeis.org

0, 1, 3, 4, 5, 6, 7, 9, 12, 13, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 33, 36, 37, 39, 45, 48, 49, 51, 52, 53, 54, 55, 57, 60, 61, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91

Offset: 1

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n)+o(n) = A000120(n), the binary weight of n. For e(n) = o(n) see A039004.

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

Cf. A000120, A139351, A139352, A139353, A139354, A139355, A039004, A139370, A139371, A139373.

Fortran
```
c See link in A139351
```

q[n_] := Module[{d = Reverse[IntegerDigits[n, 2]]}, Total@ d[[1;; -1;; 2]] >= Total@ d[[2;; -1;; 2]]]; Select[Range[0, 100], q] (* Amiram Eldar, Aug 31 2023 *)

A139373 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence lists n such that e(n) > o(n).

Original entry on oeis.org

1, 4, 5, 7, 13, 16, 17, 19, 20, 21, 22, 23, 25, 28, 29, 31, 37, 49, 52, 53, 55, 61, 64, 65, 67, 68, 69, 70, 71, 73, 76, 77, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 97, 100, 101, 103, 109, 112, 113, 115, 116, 117, 118, 119, 121, 124

Offset: 1

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

nonn

e(n)+o(n) = A000120(n), the binary weight of n. For e(n) = o(n) see A039004.

Cf. A000120, A139351, A139352, A139353, A139354, A139355.

Cf. A039004, A139370, A139371, A139372.

Fortran
```
c See link in A139351
```

aQ[n_] := Module[{d = Reverse[IntegerDigits[n,2]]}, Total@d[[1;;-1;;2]] > Total@d[[2;;-1;;2]]]; Select[Range[180], aQ] (* Amiram Eldar, Dec 15 2018 *)

isok(n) = {my(irb = Vec(select(x->(x%2), Vecrev(binary(n)), 1))); #select(x->(x%2), irb) > #irb/2;} \\ Michel Marcus, Dec 15 2018

cp's OEIS Frontend

A152715 Primes in A065049 which are not in A139370.

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A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

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A139351 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives e(n).

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A139352 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives o(n).

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A139353 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives e(n)*o(n).

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A139354 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives min{e(n), o(n)}.

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Mathematica

Formula

A139355 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives max{e(n), o(n)}.