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

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

Offset: 1

Author

Alex Ratushnyak, Jan 06 2019

Keywords

• nonn
• base

Comments

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)

Links

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

Crossrefs

Cf. A035327.

Programs

Python

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=', ')

Python

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

Extensions

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

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