A035327 -id:A035327 - cp's OEIS frontend

A096115 If n = (2^k)-1, a(n) = a((n+1)/2) = k, if n = 2^k, a(n) = a(n-1)+1 = k+1, otherwise a(n) = (A000523(n)+1)*a(A035327(n-1)).

1, 2, 2, 3, 6, 6, 3, 4, 12, 24, 24, 12, 8, 8, 4, 5, 20, 40, 40, 60, 120, 120, 60, 20, 15, 30, 30, 15, 10, 10, 5, 6, 30, 60, 60, 90, 180, 180, 90, 120, 360, 720, 720, 360, 240, 240, 120, 30, 24, 48, 48, 72, 144, 144, 72, 24, 18, 36, 36, 18, 12, 12, 6, 7, 42, 84, 84, 126

Offset: 1

Amarnath Murthy, Jun 30 2004

nonn

A fractal sequence. For k in range [1,(2^n)-1], a(2^n + k)/a(2^n - k) = n+1. Each n > 1 occurs 2*A045778(n) times in the sequence.

Permutation of A096111, i.e. a(n) = A096111(A122199(n)-1) [Note the different starting offsets]. Cf. A096113, A052330, A096114, A096116.

(define (A096115 n) (cond ((pow2? (+ n 1)) (+ 1 (A000523 n))) ((pow2? n) (+ 1 (A096115 (- n 1)))) (else (* (+ (A000523 n) 1) (A096115 (A035327 (- n 1)))))))
(define (pow2? n) (and (> n 0) (zero? (A004198bi n (- n 1)))))

Edited, extended and Scheme code added by Antti Karttunen, Aug 25 2006

A087734 a(n) = f(f(n)), where f() = A035327().

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 2, 3, 0, 1, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 0, 1, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 0, 1, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17

Offset: 0

N. J. A. Sloane, Oct 01 2003

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..16384
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.

a:= n-> ((i->Bits[Nand](i$2))@@2)(n):
seq(a(n), n=0..100);  # Alois P. Heinz, Sep 29 2019

{0}~Join~Array[Nest[BitXor[#, 2^IntegerPart[Log2@ # + 1] - 1] &, #, 2] /. -1 -> 0 &, 81] (* Michael De Vlieger, Sep 29 2019 *)

From Mikhail Kurkov, Sep 29 2019: (Start)
Some conjectures:
a(n) = n - Sum_{k=A063250(n)..A000523(n)} 2^k = n - 2^(A000523(n)+1) + 2^A063250(n) for n>0 with a(0)=0.
G.f.: 1/(1-x) * Sum_{j>=0} (2^j)*((x^(2^j))/(1+x^(2^j)) - (1-x^(2^j)) * Sum_{k>=1} x^((2^j)*(2^k-1))).
a(n) = 2*a(floor(n/2)) + n mod 2 - A036987(n) for n>1 with a(0)=a(1)=0.
a(n) = (1 - A036987(n-1))*(1 + A063250(n) - A063250(n-1))*(1 + a(n-1)) for n>0 with a(0)=0. (End)

A096116 a(1)=1, if n=(2^k)+1, a(n) = k+2, otherwise a(n) = 2+A000523(n-1)+a(2+A035327(n-1)).

Original entry on oeis.org

1, 2, 3, 5, 4, 9, 7, 6, 5, 11, 12, 14, 9, 10, 8, 7, 6, 13, 14, 16, 15, 20, 18, 17, 11, 12, 13, 15, 10, 11, 9, 8, 7, 15, 16, 18, 17, 22, 20, 19, 18, 24, 25, 27, 22, 23, 21, 20, 13, 14, 15, 17, 16, 21, 19, 18, 12, 13, 14, 16, 11, 12, 10, 9, 8, 17, 18, 20, 19, 24, 22, 21, 20, 26

Offset: 1

Amarnath Murthy, Jun 30 2004

nonn

Each n > 1 occurs A025147(n) times in the sequence.

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A096111, A050029, A050030, A052330, A096113, A096114, A096115.

a = {1}; Do[AppendTo[a, If[BitAnd[n - 1, n - 2] == 0, Log2[n - 1] + 2, 2 + Floor[Log2[n - 1]] + a[[2 + BitXor[n - 1, 2^Ceiling[Log2[n]] - 1]]]]], {n, 2, 74}]; a (* Ivan Neretin, Jun 24 2016 *)

(define (A096116 n) (cond ((= 1 n) 1) ((pow2? (- n 1)) (+ 2 (A000523 (- n 1)))) (else (+ 2 (A000523 (- n 1)) (A096116 (+ 2 (A035327 (- n 1))))))))
(define (pow2? n) (and (> n 0) (zero? (A004198bi n (- n 1)))))
;; Antti Karttunen, Aug 25 2006

Edited and extended by Antti Karttunen, Aug 25 2006

A361989 a(n) is the sum of the Fibonacci numbers missing from the dual Zeckendorf representation of n; a(0) = 0, and for n > 0, a(n) = A022290(A035327(A003754(n+1))).

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 0, 4, 3, 2, 1, 0, 7, 6, 5, 4, 3, 2, 1, 0, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0

Offset: 0

Rémy Sigrist, Apr 02 2023

We consider that a Fibonacci number is missing from the dual Zeckendorf representation of a number if it does not appear in this representation and a larger Fibonacci number appears in it.
The dual Zeckendorf representation is also known as the lazy Fibonacci representation (see A356771 for further details).
This sequence can also be seen as an irregular table T(n, k), n > 0, k = 1..A000045(n), where T(n, k) = A000045(n) - k.
a(n-1) for n>=1 is the starting position of the first occurrence of one of the longest words w in the Fibonacci word A003849 such that no length-n factor of w is repeated. The length of such words is 2n. (See links) - Gandhar Joshi, Mar 19 2024

			For n = 42:
- using F(k) = A000045(k),
- the dual Zeckendorf representation of 42 is F(8) + F(7) + F(5) + F(3) + F(2),
- the numbers F(6) and F(4) are missing,
- so a(42) = F(6) + F(4) = 8 + 3 = 11.
.
As an irregular triangle the sequence begins:
     0;
     0;
     1,  0;
     2,  1,  0;
     4,  3,  2, 1, 0;
     7,  6,  5, 4, 3, 2, 1, 0;
    12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0;
    ...

Gandhar Joshi, Walnut code and details
Index entries for sequences related to Zeckendorf expansion of n

Cf. A000045, A003754, A022290, A035327, A072649, A130312, A132665, A194029, A356771.

for (n = 1, 9, for (k = 1, f = fibonacci(n), print1 (f-k", ")))

a(n) = A000045(A072649(n)) - A194029(n) for n > 0.

a(n) = A130312(n) - A194029(n) for n > 0.

A253608 The binary representation of a(n) is the concatenation of n and the binary complement of n, A035327(n).

Original entry on oeis.org

2, 9, 12, 35, 42, 49, 56, 135, 150, 165, 180, 195, 210, 225, 240, 527, 558, 589, 620, 651, 682, 713, 744, 775, 806, 837, 868, 899, 930, 961, 992, 2079, 2142, 2205, 2268, 2331, 2394, 2457, 2520, 2583, 2646, 2709, 2772, 2835, 2898, 2961, 3024, 3087, 3150, 3213

Offset: 1

Alex Ratushnyak, Jan 05 2015

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

Cf. A035327, A036044, A070939.

a:= n-> (n+1)*(2^(ilog2(n)+1)-1):
seq(a(n), n=1..50);  # Alois P. Heinz, Jan 08 2015

Array[(# + 1) (2^(Floor@ Log2[#] + 1) - 1) &, 50] (* Michael De Vlieger, Oct 13 2018 *)

a(n) = (n+1)*(2^#binary(n)-1); \\ Michel Marcus, Jan 08 2015

for n in range(1,333):
  print(str((n+1)*(2 ** int.bit_length(int(n))-1)), end=',')

a(n) = (n+1) * (2^BL(n) - 1), where BL(n) is the binary length of n.

A323192 Numbers x such that the binary complement of x is larger than the binary complement of x^2: A035327(x) > A035327(x^2).

Original entry on oeis.org

181, 362, 724, 1448, 741455, 11863283, 99516432383215, 1592262918131443, 12738103345051545, 101904826760412361, 203809653520824722, 407619307041649444, 3260954456333195553, 6521908912666391106, 417402170410649030795, 1709679290002018430137083, 3419358580004036860274166

Offset: 1

Alex Ratushnyak, Jan 06 2019

Up to a(13) all the terms are of the form floor(sqrt(2^k-1)) for some k. - Giovanni Resta, Jan 07 2019
From Chai Wah Wu, Jan 10 2019: (Start)
The terms are numbers of the form floor(sqrt(2^(2k-1))) that are larger than 1/2 + sqrt(1/4 + 2^(2k-1) - 2^k) for some k > 0.
Theorem: if x is a term, then it is of the form floor(sqrt(2^(2k-1))) for some k > 0. In addition, floor(sqrt(2^(2k-1))) is a term if and only if it is larger than 1/2 + sqrt(1/4 + 2^(2k-1) - 2^k).
Proof: suppose x has k bits in its binary representation, i.e., 2^(k-1) <= x < 2^k. Then x^2 has either 2k-1 or 2k bits.
First we show that if x is a term of the sequence, then x^2 has 2k-1 bits. Suppose x^2 has 2k bits, i.e., x^2 >= 2^(2k-1). Then 2^2k - 1 - x^2 < 2^k - 1 - x. This is rearranged as x^2 - x + 2^k - 2^2k > 0. Solving this quadratic inequality leads to x > 2^k which contradicts the fact that x has k bits.
Thus x^2 < 2^(2k-1) and 2^(2k-1) - 1 - x^2 < 2^k - 1 - x. Solving this inequality and combining it with x < sqrt(2^(2k-1)) shows that x must satisfy sqrt(2^(2k-1)) > x > 1/2 + sqrt(1/4 + 2^(2k-1) - 2^k).
To complete the proof, we need to show that for each k, there is at most one integer satisfying this inequality. This is easily verified for k = 1. Assume that k > 1. Let a = sqrt(2^(2k-1)) and b = 1/2 + sqrt(1/4 + 2^(2k-1) - 2^k).
Using the identity sqrt(x) - sqrt(y) = (x-y)/(sqrt(x)+sqrt(y)) it follows that a-b = -1/2 + (2^k - 1/4)/(sqrt(2^(2k-1))+sqrt(1/4 + 2^(2k-1) - 2^k) < -1/2 + 2^k/(sqrt(2^(2k-1)) + sqrt(2^(2k-1)-2^k)).
Since k >= 2, sqrt(2^(2k-1)-2^k) = sqrt(2^(2k-1)(1-2^(1-k))) >= sqrt(1/2)*sqrt(2^(2k-1)). This implies that a-b < -1/2 + 2^k/((1+sqrt(1/2))*(sqrt(2^(2k-1)))) = 2*sqrt(2)/(sqrt(2)+1) - 1/2 < 0.672. This implies that there is at most one integer between b and a.
The above discussion also provides another characterization of the sequence. floor(sqrt(2^(2k-1))) is a term if and only if sqrt(2^(2k-1))-floor(sqrt(2^(2k-1))) < a-b where a-b is as defined above.
The criterion can be simplified as:
floor(sqrt(2^(2k-1))) is a term if and only if it is larger than 1/2 + sqrt(1/4 + 2^(2k-1) - 2^k). This concludes the proof.
Note that a-b -> 0.5 as k -> oo, i.e., for large k, the fractional part of sqrt(2^(2k-1)) should be less than about 0.5 in order for the integer part to be a term.
(End)

Chai Wah Wu, Table of n, a(n) for n = 1..721

Cf. A035327.

for n in range(1,20000000):
    b1 = (1 << n.bit_length()) - 1
    n2 = n*n
    b2 = (1 << n2.bit_length()) - 1
    if b1-n > b2-n2:  print(str(n), end=', ')

from sympy import integer_nthroot
A323192_list = []
for k in range(1000):
    n = integer_nthroot(2**(2*k+1),2)[0]
    if n*(n-1) + 2**(len(bin(n))-2) - 2**(len(bin(n**2))-2) > 0:
        A323192_list.append(n) # Chai Wah Wu, Jan 10 2019

a(7)-a(13) from Giovanni Resta, Jan 07 2019

a(14)-a(17) from Chai Wah Wu, Jan 10 2019

A323066 Numbers whose binary complement (A035327) is a square.

Original entry on oeis.org

0, 1, 2, 3, 6, 7, 11, 14, 15, 22, 27, 30, 31, 38, 47, 54, 59, 62, 63, 78, 91, 102, 111, 118, 123, 126, 127, 134, 155, 174, 191, 206, 219, 230, 239, 246, 251, 254, 255, 286, 315, 342, 367, 390, 411, 430, 447, 462, 475, 486, 495, 502, 507, 510, 511, 539, 582, 623, 662

Offset: 1

Alex Ratushnyak, Jan 03 2019

The binary complement of every square is in the sequence. Using that might ease computation. - David A. Corneth, Jan 08 2019

			The binary complement of 22 is 9. Because 9 is a square, 22 is in the sequence.
The binary complement of 4^2 = 16 is 15 so 15 is in the sequence. - _David A. Corneth_, Jan 08 2019

Cf. A035327, A000290, A000225 is a subsequence.

q:= n-> issqr(Bits[Nand](n$2)):
select(q, [$0..1000])[];  # Alois P. Heinz, Sep 03 2021

Select[Range@ 700, IntegerQ@ Sqrt@ FromDigits[IntegerDigits[#, 2] /. {0 -> 1, 1 -> 0}, 2] &] (* Michael De Vlieger, Jan 04 2019 *)

bc(n) = bitxor(n, 2^(1+logint(max(n, 1), 2))-1); \\ A035327
isok(n) = issquare(bc(n)); \\ Michel Marcus, Jan 04 2019

A348074 a(n) = A035327(n) if n is even, 3n + 1 if n is odd.

Original entry on oeis.org

1, 4, 1, 10, 3, 16, 1, 22, 7, 28, 5, 34, 3, 40, 1, 46, 15, 52, 13, 58, 11, 64, 9, 70, 7, 76, 5, 82, 3, 88, 1, 94, 31, 100, 29, 106, 27, 112, 25, 118, 23, 124, 21, 130, 19, 136, 17, 142, 15, 148, 13, 154, 11, 160, 9, 166, 7, 172, 5, 178, 3, 184, 1, 190, 63

Offset: 0

Alexander Van Plantinga, Sep 27 2021

nonn

			n    parity   function       a(n)
=    =====    =========     ======
0     even    A035327(0)      1
1     odd      3*1 + 1        4
2     even    A035327(2)      1
3     odd      3*3 + 1       10
4     even    A035327(4)      3
5     odd      3*5 + 1       16

Cf. A006370, A035327.

def A348074(n):
    if n%2==0:
        if n==0: return 1
        else: return (1<A035327(n)
    else: return 3*n+1

def A348074(n): return 3*n+1 if n&1 or n==0 else (~n)^(-1<Chai Wah Wu, Dec 20 2022

A323064 Squares whose binary complement (A035327) is a cube.

Original entry on oeis.org

0, 1, 36, 100, 484, 131044

Offset: 1

Alex Ratushnyak, Jan 03 2019

Square roots: 0, 1, 6, 10, 22, 362.

			The binary complement of 100 is 27. Because 27 is a cube, 100 is in the sequence.

Cf. A035327, A000290, A000578.

isok(n) = n == 0 || (issquare(n) && ispower(2^(1+logint(n, 2))-1-n, 3))
for (n=0, 500, if (isok(n^2), print1(n^2, ", "))); \\ Michel Marcus, Jan 04 2019

A323067 Primes whose binary complement (A035327) is a square.

Original entry on oeis.org

2, 3, 7, 11, 31, 47, 59, 127, 191, 239, 251, 367, 827, 1019, 1471, 1723, 2011, 2939, 4079, 4091, 4591, 6427, 7867, 8191, 10607, 11483, 12539, 13679, 13883, 14447, 14783, 14939, 15227, 15359, 16127, 16187, 16319, 18367, 26683, 27583, 28411, 29167, 29851, 32191

Offset: 1

Alex Ratushnyak, Jan 03 2019

Cf. A000040, A035327, A000290, A323066.

Select[Prime@ Range[10^4], IntegerQ@ Sqrt@ FromDigits[IntegerDigits[#, 2] /. {0 -> 1, 1 -> 0}, 2] &] (* Michael De Vlieger, Jan 04 2019 *)

bc(n) = bitxor(n, 2^(1+logint(max(n, 1), 2))-1); \\ A035327
isok(n) = isprime(n) && issquare(bc(n)); \\ Michel Marcus, Jan 04 2019

cp's OEIS Frontend

A096115 If n = (2^k)-1, a(n) = a((n+1)/2) = k, if n = 2^k, a(n) = a(n-1)+1 = k+1, otherwise a(n) = (A000523(n)+1)*a(A035327(n-1)).

Views

Author

Keywords

Comments

Crossrefs

Programs

Scheme

Extensions

A087734 a(n) = f(f(n)), where f() = A035327().

Views

Author

Keywords

Links

Programs

Maple

Mathematica

Formula

A096116 a(1)=1, if n=(2^k)+1, a(n) = k+2, otherwise a(n) = 2+A000523(n-1)+a(2+A035327(n-1)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Scheme

Extensions

A361989 a(n) is the sum of the Fibonacci numbers missing from the dual Zeckendorf representation of n; a(0) = 0, and for n > 0, a(n) = A022290(A035327(A003754(n+1))).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A253608 The binary representation of a(n) is the concatenation of n and the binary complement of n, A035327(n).

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

A323192 Numbers x such that the binary complement of x is larger than the binary complement of x^2: A035327(x) > A035327(x^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Python

Python

Extensions

A323066 Numbers whose binary complement (A035327) is a square.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

A348074 a(n) = A035327(n) if n is even, 3n + 1 if n is odd.

Views

Author

Keywords

Examples