A063694 -id:A063694 - cp's OEIS frontend

A062880 Zero together with the numbers which can be written as a sum of distinct odd powers of 2.

0, 2, 8, 10, 32, 34, 40, 42, 128, 130, 136, 138, 160, 162, 168, 170, 512, 514, 520, 522, 544, 546, 552, 554, 640, 642, 648, 650, 672, 674, 680, 682, 2048, 2050, 2056, 2058, 2080, 2082, 2088, 2090, 2176, 2178, 2184, 2186, 2208, 2210, 2216, 2218, 2560, 2562

Offset: 0

Antti Karttunen, Jun 26 2001

Binary expansion of n does not contain 1-bits at even positions.
Integers whose base-4 representation consists of only 0's and 2's.
Every nonnegative even number is a unique sum of the form a(k)+2*a(l); moreover, this sequence is unique with such property. - Vladimir Shevelev, Nov 07 2008
Also numbers such that the digital sum base 2 and the digital sum base 4 are in a ratio of 2:4. - Michel Marcus, Sep 23 2013
From Gus Wiseman, Jun 10 2020: (Start)
Numbers k such that the k-th composition in standard order has all even parts. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. For example, the sequence of all compositions into even parts begins:
()            520: (6,4)          2080: (6,6)
(2)           522: (6,2,2)        2082: (6,4,2)
(4)           544: (4,6)          2088: (6,2,4)
(2,2)         546: (4,4,2)        2090: (6,2,2,2)
(6)           552: (4,2,4)        2176: (4,8)
(4,2)         554: (4,2,2,2)      2178: (4,6,2)
(2,4)         640: (2,8)          2184: (4,4,4)
(2,2,2)       642: (2,6,2)        2186: (4,4,2,2)
(8)           648: (2,4,4)        2208: (4,2,6)
(6,2)         650: (2,4,2,2)      2210: (4,2,4,2)
(4,4)         672: (2,2,6)        2216: (4,2,2,4)
(4,2,2)       674: (2,2,4,2)      2218: (4,2,2,2,2)
(2,6)         680: (2,2,2,4)      2560: (2,10)
(2,4,2)       682: (2,2,2,2,2)    2562: (2,8,2)
(2,2,4)      2048: (12)           2568: (2,6,4)
(2,2,2,2)    2050: (10,2)         2570: (2,6,2,2)
(10)         2056: (8,4)          2592: (2,4,6)
(8,2)        2058: (8,2,2)        2594: (2,4,4,2)
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
D. H. Bailey, J. M. Borwein, R. E. Crandall, and C. Pomerance, On the binary expansions of algebraic numbers, J. Théor. Nombres Bordeaux, 16 (2004), 487-518.
S. Eigen, A. Hajian, and S. Kalikow, Ergodic transformations and sequences of integers, Israel J. Math. 75 (1991), 119-128; Math. Rev. 1147294 (93c:28014).

Cf. A000695.
Except for first term, n such that A063694(n) = 0. Binary expansion is given in A062033.
Interpreted as Zeckendorf expansion: A062879.
Central diagonal of arrays A163357 and A163359.
Even partitions are counted by A035363.
Numbers with an even number of 1's in binary expansion are A001969.
Numbers whose binary expansion has even length are A053754.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Compositions without even parts are A060142.
- Sum is A070939.
- Product is A124758.
- Strict compositions are A233564.
- Heinz number is A333219.
- Number of distinct parts is A334028.

uint32_t a_next(uint32_t a_n) { return (a_n + 0x55555556) & 0xaaaaaaaa; } /* Falk Hüffner, Jan 22 2022 */

a062880 n = a062880_list !! n
a062880_list = filter f [0..] where
   f 0 = True
   f x = (m == 0 || m == 2) && f x'  where (x', m) = divMod x 4
-- Reinhard Zumkeller, Nov 20 2012

[seq(a(j),j=0..100)]; a := n -> add((floor(n/(2^i)) mod 2)*(2^((2*i)+1)),i=0..floor_log_2(n+1));

b[n_] := BitAnd[n, Sum[2^k, {k, 0, Log[2, n] // Floor, 2}]]; Select[Range[ 0, 10^4], b[#] == 0&] (* Jean-François Alcover, Feb 28 2016 *)

def A062880(n): return int(bin(n)[2:],4)<<1 # Chai Wah Wu, Aug 21 2023

a(n) = 2 * A000695(n). - Vladimir Shevelev, Nov 07 2008
From Robert Israel, Apr 10 2018: (Start)
a(2*n) = 4*a(n).
a(2*n+1) = 4*a(n)+2.
G.f. g(x) satisfies: g(x) = 4*(1+x)*g(x^2)+2*x/(1-x^2). (End)

A063695 Remove even-positioned bits from the binary expansion of n.

Original entry on oeis.org

0, 0, 2, 2, 0, 0, 2, 2, 8, 8, 10, 10, 8, 8, 10, 10, 0, 0, 2, 2, 0, 0, 2, 2, 8, 8, 10, 10, 8, 8, 10, 10, 32, 32, 34, 34, 32, 32, 34, 34, 40, 40, 42, 42, 40, 40, 42, 42, 32, 32, 34, 34, 32, 32, 34, 34, 40, 40, 42, 42, 40, 40, 42, 42, 0, 0, 2, 2, 0, 0, 2, 2, 8, 8, 10, 10, 8, 8, 10, 10, 0, 0

Offset: 0

Antti Karttunen, Aug 03 2001

In the base 4 expansion of n, change 1 to 0 and 3 to 2. - Paolo Xausa, Feb 27 2025

			a(25) = 8 because 25 = 11001 in binary and when we AND this with 1010 we are left with 1000 = 8.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions

Cf. A004514 (bisection), A063694 (remove odd-positioned bits), A090569.

a063695 0 = 0
a063695 n = 4 * a063695 n' + 2 * div q 2
            where (n', q) = divMod n 4
-- Reinhard Zumkeller, Sep 26 2015

[seq(every_other_pos(j,2,1),j=0..120)]; # Function every_other_pos given at A063694.

A063695[n_] := FromDigits[ReplaceAll[IntegerDigits[n, 4], {1 -> 0, 3 -> 2}], 4];
Array[A063695, 100, 0] (* Paolo Xausa, Feb 27 2025 *)

def A063695(n): return n&((1<<(m:=n.bit_length())+(m&1^1))-1)//3 # Chai Wah Wu, Jan 30 2023

a(n) + A063694(n) = n.
a(n) = 2*(floor(n/2)-a(floor(n/2))). - Vladeta Jovovic, Feb 23 2003
From Ralf Stephan, Oct 06 2003: (Start)
G.f. 1/(1-x) * Sum_{k>=0} (-2)^k*2t^2/(1-t^2) where t = x^2^k.
Members of A004514 written twice.
(End)
a(n) = 4 * a(floor(n / 4)) + 2 * floor(n mod 4 / 2). - Reinhard Zumkeller, Sep 26 2015
a(n) = A090569(n+1)-1. - R. J. Mathar, Jun 22 2020
a(n) = 2*(n - A380110(n)). - Paolo Xausa, Feb 27 2025

A088442 A linear version of the Josephus problem.

Original entry on oeis.org

1, 3, 1, 3, 9, 11, 9, 11, 1, 3, 1, 3, 9, 11, 9, 11, 33, 35, 33, 35, 41, 43, 41, 43, 33, 35, 33, 35, 41, 43, 41, 43, 1, 3, 1, 3, 9, 11, 9, 11, 1, 3, 1, 3, 9, 11, 9, 11, 33, 35, 33, 35, 41, 43, 41, 43, 33, 35, 33, 35, 41, 43, 41, 43, 129, 131, 129, 131, 137, 139, 137, 139, 129, 131

Offset: 0

N. J. A. Sloane, Nov 09 2003

Or a(n) is in A145812 such that (2*n + 3 - a(n))/2 is in A145812 as well. Note also that a(n) + 2*A090569(n+1) = 2*n + 3. - Vladimir Shevelev, Oct 20 2008

			If n=4, 2n+1 = 9 = 1 + 0*2 + 0*2^2 + 1*2^3, so a(4) = 1 + 0*2 + 1*2^3 = 9.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
C. Groer, The Mathematics of Survival: From Antiquity to the Playground, Amer. Math. Monthly, 110 (No. 9, 2003), 812-825. (This is the sequence W(2n+1).)
Index entries for sequences related to the Josephus Problem

Cf. A006257, A063694, A088443, A088452, A090569, A145812.

a088442 = (+ 1) . a004514  -- Reinhard Zumkeller, Sep 26 2015

function A063694(n)
  if n le 1 then return n;
  else return 4*A063694(Floor(n/4)) + (n mod 2);
  end if; return A063694;
end function;
A088442:= func< n | 2*A063694(n) + 1 >;
[A088442(n): n in [0..100]]; // G. C. Greubel, Dec 05 2022

a:=proc(n) local b: b:=convert(2*n+1,base,2): 1+sum(b[2*j]*2^(2*j-1),j=1..nops(b)/2) end: seq(a(n),n=0..100);
with(Bits): seq(And(2*n+1, convert("aaaaaa", decimal, hex)) + 1, n=0..127); # Georg Fischer, Dec 03 2022

A004514[n_]:= A004514[n]= If[n==0, 0, 2*(n-A004514[Floor[n/2]])];
A088442[n_] := A004514[n] +1;
Table[A088442[n], {n,0,100}] (* G. C. Greubel, Dec 05 2022 *)

def A088442(n): return ((n&((1<<(m:=n.bit_length())+(m&1))-1)//3)<<1)+1 # Chai Wah Wu, Jan 30 2023

def A063694(n):
    if (n<2): return n
    else: return 4*A063694(floor(n/4)) + (n%2)
def A088442(n): return 2*A063694(n) + 1
[A088442(n) for n in range(101)] # G. C. Greubel, Dec 05 2022

To get a(n), write 2n+1 as Sum b_j 2^j, then a(n) = 1 + Sum_{j odd} b_j 2^j.
Equals A004514(n) + 1. - Chris Groer (cgroer(AT)math.uga.edu), Nov 10 2003
a(n) = 2*A063694(n) + 1. - G. C. Greubel, Dec 05 2022

More terms from Emeric Deutsch, May 27 2004

A063698 Positions of negative coefficients in cyclotomic polynomial Phi_n(x), converted from binary to decimal. (The constant term in the least significant bit (bit-0), the term x in the next bit (bit-1) and so on).

Original entry on oeis.org

0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 10, 0, 4, 0, 42, 146, 0, 0, 8, 0, 68, 2322, 682, 0, 16, 0, 2730, 0, 1092, 0, 56, 0, 0, 599186, 43690, 8726850, 64, 0, 174762, 9585810, 4112, 0, 792, 0, 279620, 2101256, 2796202, 0, 256, 0, 32800, 2454267026, 4473924, 0, 512

Offset: 0

Antti Karttunen, Aug 03 2001

nonn

Maple procedures Phi_pos_terms and Phi_neg_terms are modeled after the formula given in Lam and Leung paper and they compute correct results for all integers x > 1 and for all n with at most two distinct odd prime factors (that is, up to n=104). Other procedures as in A063696 and A063694.

Antti Karttunen, Table of n, a(n) for n = 0..3003
D. M. Bloom, On the Coefficients of the Cyclotomic Polynomials, Amer.Math.Monthly 75, 372-377, 1968.
T.Y. Lam and K. H. Leung, On the Cyclotomic Polynomial Phi_pq(X), Amer.Math.Monthly 103, 562-564, August-September 1996.
H. Lenstra, Vanishing sums of roots of unity, in Proc. Bicentennial Congress Wiskundig Genootschap (Vrije Univ. Amsterdam, 1978), Part II, pp. 249-268.
Index entries for cyclotomic polynomials, positions of coefficients

A013594, A063696 gives the positions of the positive and A063697 the nonzero terms. This sequence in binary: A063699. A019320[n] = A063696[n]-A063698[n] for up to n=104

with(numtheory); [seq(Phi_neg_terms(j,2),j=0..104)];
Phi_neg_terms := proc(n,x) local a,m,p,q,e,f,r,s; if(n < 2) then RETURN(n); fi; a := op(2, ifactors(n)); m := nops(a); p := a[1][1]; e := a[1][2]; if(1 = m) then RETURN(0); fi; if(2 = m) then q := a[2][1]; f := a[2][2]; r := inv_p_mod_q(p,q)-1; s := inv_p_mod_q(q,p)-1;
RETURN( x^((s+1)*(q^f)*(p^(e-1))) * x^((r+1)*(p^e)*(q^(f-1))) * x^(-((p^e) * (q^f))) * (`if`((p-2)=s,1,(((x^((p-s-1)*((q^f)*(p^(e-1)))))-1)/((x^((q^f)*(p^(e-1))))-1)))) * (`if`((q-2)=r,1,(((x^((q-r-1)*((p^e)*(q^(f-1)))))-1)/((x^((p^e)*(q^(f-1))))-1)))) ); fi;
if((3 = m) and (2 = p)) then if(1 = e) then RETURN(every_other_pos(Phi_neg_terms(n/2,x),x,0)+every_other_pos(Phi_pos_terms(n/2,x),x,1)); else RETURN(dilate(Phi_neg_terms((n/(2^(e-1))),x),x,2^(e-1))); fi; else printf(`Cannot handle argument %a with >=3 distinct odd prime factors!\n`,n); RETURN(0); fi; end;

a[n_] := 2^(Flatten[Position[CoefficientList[Cyclotomic[n, x], x], ?Negative]] - 1) // Total; a[0] = 0; Table[a[n], {n, 0, 60}] (* _Jean-François Alcover, Mar 05 2016 *)

a(n)=my(p); if(n<1, 0, p=polcyclo(n); sum(i=0, n, 2^i*(polcoeff(p, i)<0))) \\ Michel Marcus, Mar 05 2016

A366244 The largest infinitary divisor of n that is a term of A366242.

Original entry on oeis.org

1, 2, 3, 1, 5, 6, 7, 2, 1, 10, 11, 3, 13, 14, 15, 16, 17, 2, 19, 5, 21, 22, 23, 6, 1, 26, 3, 7, 29, 30, 31, 32, 33, 34, 35, 1, 37, 38, 39, 10, 41, 42, 43, 11, 5, 46, 47, 48, 1, 2, 51, 13, 53, 6, 55, 14, 57, 58, 59, 15, 61, 62, 7, 16, 65, 66, 67, 17, 69, 70, 71

Offset: 1

Amiram Eldar, Oct 05 2023

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

Cf. A063694, A366242, A366243, A366245.

See the formula section for the relationships with A007913, A046100, A059895, A059896, A059897, A225546, A247503, A352780.

f[p_, e_] := p^BitAnd[e, Sum[2^k, {k, 0, Floor@ Log2[e], 2}]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

s(e) = sum(k = 0, e, (-2)^k*floor(e/2^k));
a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^s(f[i,2]));}

Multiplicative with a(p^e) = p^A063694(e).
a(n) = n / A366245(n).
a(n) >= 1, with equality if and only if n is a term of A366243.
a(n) <= n, with equality if and only if n is a term of A366242.
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} (1-1/p)*(Sum_{k>=1} p^(A063694(k)-2*k)) = 0.35319488024808595542... .
From Peter Munn, Jan 09 2025: (Start)
a(n) = max({k in A366242 : A059895(k, n) = k}).
a(n) = Product_{k >= 0} A352780(n, 2k).
Also defined by:
- for n in A046100, a(n) = A007913(n);
- a(n^4) = (a(n))^4;
- a(A059896(n,k)) = A059896(a(n), a(k)).
Other identities:
a(n) = sqrt(A366245(n^2)).
a(A059897(n,k)) = A059897(a(n), a(k)).
a(A225546(n)) = A225546(A247503(n)).
(End)

A004514 Generalized nim sum n + n in base 4.

Original entry on oeis.org

0, 2, 0, 2, 8, 10, 8, 10, 0, 2, 0, 2, 8, 10, 8, 10, 32, 34, 32, 34, 40, 42, 40, 42, 32, 34, 32, 34, 40, 42, 40, 42, 0, 2, 0, 2, 8, 10, 8, 10, 0, 2, 0, 2, 8, 10, 8, 10, 32, 34, 32, 34, 40, 42, 40, 42, 32, 34, 32, 34, 40, 42, 40, 42, 128, 130, 128, 130, 136, 138, 136, 138, 128

Offset: 0

N. J. A. Sloane

In the base 4 expansion of 2*n + 1, change 1 to 0 and 3 to 2. - Paolo Xausa, Feb 27 2025

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for sequences related to Nim-sums

Cf. A053985, A063694, A063695, A088442.

a004514 = a063695 . (+ 1) . (* 2)  -- Reinhard Zumkeller, Sep 26 2015

function A063694(n)
  if n le 1 then return n;
  else return 4*A063694(Floor(n/4)) + ((n mod 4) mod 2);
  end if; return A063694;
end function;
A004514:= func< n | 2*A063694(n) >;
[A004514(n): n in [0..90]]; // G. C. Greubel, Dec 05 2022

A004514[n_]:= A004514[n]= If[n==0, 0, 2*n -2*A004514[Floor[n/2]]];
Table[A004514[n], {n,0,90}] (* G. C. Greubel, Dec 05 2022 *)
A004514[n_] := FromDigits[ReplaceAll[IntegerDigits[2*n + 1, 4], {1 -> 0, 3 -> 2}], 4];
Array[A004514, 100, 0] (* Paolo Xausa, Feb 27 2025 *)

a(n) = if(n, bitand(n, 2<Kevin Ryde, Dec 10 2022

def A004514(n): return (n&((1<<(m:=n.bit_length())+(m&1))-1)//3)<<1 # Chai Wah Wu, Jan 30 2023

def A063694(n):
    if (n<2): return n
    else: return 4*A063694(floor(n/4)) + ((n%4)%2)
def A004514(n): return 2*A063694(n)
[A004514(n) for n in range(91)] # G. C. Greubel, Dec 05 2022

Generalized nim sum m + n in base q: write m and n in base q and add mod q with no carries, e.g., 5 + 8 in base 3 = "21" + "22" = "10" = 1.
From Vladeta Jovovic, Feb 23 2003: (Start)
a(n) = 2*(n - a(floor(n/2))).
a(n) = 2*A063694(n). (End)
a(n) = A088442(n) - 1. - Chris Groer (cgroer(AT)math.uga.edu), Nov 10 2003
a(n) = n + A053985(n). - Reinhard Zumkeller, Dec 27 2003
a(n) = A063695(2*n+1). - Reinhard Zumkeller, Sep 26 2015
a(n) = Sum_{k>=0} A030308(n,k)*A103424(k+1). - Philippe Deléham, Jan 12 2023

More terms from Reinhard Zumkeller, Dec 27 2003

A063696 Positions of positive coefficients in cyclotomic polynomial Phi_n(x), converted from binary to decimal.

Original entry on oeis.org

0, 2, 3, 7, 5, 31, 5, 127, 17, 73, 21, 2047, 17, 8191, 85, 297, 257, 131071, 65, 524287, 273, 4681, 1365, 8388607, 257, 1082401, 5461, 262657, 4369, 536870911, 387, 2147483647, 65537, 1198665, 87381, 17454241, 4097, 137438953471, 349525

Offset: 0

Antti Karttunen, Aug 03 2001

nonn

Maple procedures Phi_pos_terms and Phi_neg_terms are modeled after the formula given in Lam and Leung paper and they compute correct results for all integers x > 1 and for all n with at most two distinct odd prime factors (that is, up to n=104). Other procedures as in A063698 and A063694.

D. M. Bloom, On the Coefficients of the Cyclotomic Polynomials, Amer.Math.Monthly 75, 372-377, 1968.
T. Y. Lam and K. H. Leung, On the Cyclotomic Polynomial Phi_pq(X), Amer.Math.Monthly 103, 562-564, August-September 1996.
H. W. Lenstra, Vanishing sums of roots of unity, in Proc. Bicentennial Congress Wiskundig Genootschap (Vrije Univ. Amsterdam, 1978), Part II, pp. 249-268.
Index entries for cyclotomic polynomials, positions of coefficients

Cf. A013594, A063697 (binary version), A063698 (negative terms), A063670 (nonzero terms).

A019320(n) = a(n) - A063698(n) for up to n=104.

with(numtheory); [seq(Phi_pos_terms(j,2),j=0..104)];
inv_p_mod_q := (p,q) -> op(2,op(1,msolve(p*x=1,q))); # Find's p's inverse modulo q.
dilate := proc(nn,x,e) local n,i,s; n := nn; i := 0; s := 0; while(n > 0) do s := s + (((x^e)^i)*(n mod x)); n := floor(n/x); i := i+1; od; RETURN(s); end;
Phi_pos_terms := proc(n,x) local a,m,p,q,e,f,r,s; if(n < 2) then RETURN(x); fi; a := op(2, ifactors(n)); m := nops(a); p := a[1][1]; e := a[1][2]; if(1 = m) then RETURN(((x^(p^e))-1)/((x^(p^(e-1)))-1)); fi; if(2 = m) then q := a[2][1]; f := a[2][2]; r := inv_p_mod_q(p,q)-1; s := inv_p_mod_q(q,p)-1; RETURN( (`if`(0=s,1,(((x^((s+1)*((q^f)*(p^(e-1)))))-1)/((x^((q^f)*(p^(e-1))))-1)))) * (`if`(0=r,1,(((x^((r+1)*((p^e)*(q^(f-1)))))-1)/((x^((p^e)*(q^(f-1))))-1)))) ); fi; if((3 = m) and (2 = p)) then if(1 = e) then RETURN(every_other_pos(Phi_pos_terms(n/2,x),x,0)+every_other_pos(Phi_neg_terms(n/2,x),x,1)); else RETURN(dilate(Phi_pos_terms((n/(2^(e-1))),x),x,2^(e-1))); fi; else printf(`Cannot handle argument %a with three or more distinct odd prime factors!\n`,n); RETURN(0); fi; end;

a[n_] := 2^(Flatten[Position[CoefficientList[Cyclotomic[n, x], x], ?Positive]] - 1) // Total; a[0] = 0; Table[a[n], {n, 0, 40}] (* _Jean-François Alcover, Mar 05 2016 *)

a(n)=local(p); if(n<1,0,p=polcyclo(n); sum(i=0,n,2^i*(polcoeff(p,i)>0)))

A371442 For any positive integer n with binary digits (b_1, ..., b_w) (where b_1 = 1), the binary digits of a(n) are (b_1, b_3, ..., b_{2*ceiling(w/2)-1}); a(0) = 0.

Original entry on oeis.org

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

Offset: 0

Rémy Sigrist, Mar 24 2024

In other words, we keep odd-indexed bits.

For any v > 0, the value v appears A003945(A070939(v)) times in the sequence.

			The first terms, in decimal and in binary, are:
  n   a(n)  bin(n)  bin(a(n))
  --  ----  ------  ---------
   0     0       0          0
   1     1       1          1
   2     1      10          1
   3     1      11          1
   4     2     100         10
   5     3     101         11
   6     2     110         10
   7     3     111         11
   8     2    1000         10
   9     2    1001         10
  10     3    1010         11
  11     3    1011         11
  12     2    1100         10
  13     2    1101         10
  14     3    1110         11
  15     3    1111         11

Rémy Sigrist, Table of n, a(n) for n = 0..8192
Index entries for sequences related to binary expansion of n

See A371459 for the sequence related to even-indexed bits.
See A059905 and A063694 for similar sequences.
Cf. A000695, A001196, A003945, A070939, A165199, A371461.

A371442[n_] := FromDigits[IntegerDigits[n, 2][[1;;-1;;2]], 2];
Array[A371442, 100, 0] (* Paolo Xausa, Mar 28 2024 *)

a(n) = { my (b = binary(n)); fromdigits(vector(ceil(#b/2), k, b[2*k-1]), 2); }

Python
```
def a(n): return int(bin(n)[::2], 2)
```

a(A000695(n)) = n.
a(A001196(n)) = n.
a(A165199(n)) = a(n).

A345290 a(n) is obtained by replacing 2^k in binary expansion of n with Fibonacci(-k-2).

Original entry on oeis.org

0, -1, 2, 1, -3, -4, -1, -2, 5, 4, 7, 6, 2, 1, 4, 3, -8, -9, -6, -7, -11, -12, -9, -10, -3, -4, -1, -2, -6, -7, -4, -5, 13, 12, 15, 14, 10, 9, 12, 11, 18, 17, 20, 19, 15, 14, 17, 16, 5, 4, 7, 6, 2, 1, 4, 3, 10, 9, 12, 11, 7, 6, 9, 8, -21, -22, -19, -20, -24

Offset: 0

Rémy Sigrist, Jun 13 2021

This sequence is a variant of A022290; here we consider Fibonacci numbers with negative indices (A039834), there Fibonacci numbers with positive indices (A000045).

After the initial 0, the sequence alternates runs of positive terms and runs of negative terms, the k-th run having 2^(k-1) terms.

			For n = 3:
- 3 = 2^1 + 2^0,
- so a(3) = A039834(2+1) + A039834(2+0) = 2 - 1 = 1.

Rémy Sigrist, Table of n, a(n) for n = 0..8191

Cf. A000045, A000695, A020989, A022290, A039834, A062880, A063694, A063695, A072197, A080675, A136412, A345291, A345292.

a(n) = { my (v=0, e); while (n, n-=2^e=valuation(n, 2); v+=fibonacci(-2-e)); v }

a(n) = A022290(A063695(n)) - A022290(A063694(n)).
a(n) = A022290(n) iff n belongs to A062880.
a(n) = -A022290(n) iff n belongs to A000695.
a(n) = 0 iff n = 0.
a(n) = 1 iff n belongs to A072197.
a(n) = 2 iff n belongs to A080675.
a(n) = -1 iff n belongs to A020989.
a(n) = -2 iff n belongs to A136412.

A377414 a(n) is the largest term of A126684, say b, such that n AND b = b (where AND denotes the bitwise AND operator).

Original entry on oeis.org

0, 1, 2, 2, 4, 5, 4, 5, 8, 8, 10, 10, 8, 8, 10, 10, 16, 17, 16, 17, 20, 21, 20, 21, 16, 17, 16, 17, 20, 21, 20, 21, 32, 32, 34, 34, 32, 32, 34, 34, 40, 40, 42, 42, 40, 40, 42, 42, 32, 32, 34, 34, 32, 32, 34, 34, 40, 40, 42, 42, 40, 40, 42, 42, 64, 65, 64, 65

Offset: 0

Rémy Sigrist, Oct 27 2024

For any n > 0 with binary expansion (b_1 = 1, b_2, ..., b_k), the binary expansion of a(n) is (c_1, ..., c_k) where c_i = b_i when i is odd, c_i = 0 when i is even.

For any n, the value c = n - a(n) also belongs to A126684 and satisfies n AND c = c (see A377415).

			The first terms, in decimal and in binary, are:
  n   a(n)  bin(n)  bin(a(n))
  --  ----  ------  ---------
   0     0       0          0
   1     1       1          1
   2     2      10         10
   3     2      11         10
   4     4     100        100
   5     5     101        101
   6     4     110        100
   7     5     111        101
   8     8    1000       1000
   9     8    1001       1000
  10    10    1010       1010
  11    10    1011       1010
  12     8    1100       1000
  13     8    1101       1000
  14    10    1110       1010
  15    10    1111       1010

Rémy Sigrist, Table of n, a(n) for n = 0..8191
Index entries for sequences related to binary expansion of n

See A063694, A063695 and A374356 for similar sequences.

Cf. A000975, A070939, A126684, A371442, A377415.

a(n) = { my (v = 0, x = exponent(n), y); while (n, n -= 2^y = exponent(n); if (x%2 == y%2, v += 2^y;);); return (v); }

a(n) <= n with equality iff n belongs to A126684.
a(a(n)) = a(n).
a(2*n) = 2*a(n).
a(n) = n AND A000975(A070939(n)). - Alan Michael Gómez Calderón, Jun 27 2025

cp's OEIS Frontend

A062880 Zero together with the numbers which can be written as a sum of distinct odd powers of 2.

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A063695 Remove even-positioned bits from the binary expansion of n.

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A088442 A linear version of the Josephus problem.

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A063698 Positions of negative coefficients in cyclotomic polynomial Phi_n(x), converted from binary to decimal. (The constant term in the least significant bit (bit-0), the term x in the next bit (bit-1) and so on).

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A366244 The largest infinitary divisor of n that is a term of A366242.

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A004514 Generalized nim sum n + n in base 4.

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