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Numbers whose digits in base 2 are in nonincreasing order.

Might be called "nialpdromes".

Subset of A077436. Proof: Since a(n) is of the form (2^i-1)*2^j, i,j >= 0, a(n)^2 = (2^(2i) - 2^(i+1))*2^(2j) + 2^(2j) where the first sum term has i-1 one bits and its 2j-th bit is zero, while the second sum term switches the 2j-th bit to one, giving i one bits, as in a(n). - Ralf Stephan, Mar 08 2004

Numbers whose binary representation contains no "01". - Benoit Cloitre, May 23 2004

Every polynomial with coefficients equal to 1 for the leading terms and 0 after that, evaluated at 2. For instance a(13) = x^4 + x^3 + x^2 at 2, a(14) = x^4 + x^3 + x^2 + x at 2. - Ben Paul Thurston, Jan 11 2008

From Gary W. Adamson, Jul 18 2008: (Start)

As a triangle by rows starting:

1;

2, 3;

4, 6, 7;

8, 12, 14, 15;

16, 24, 28, 30, 31;

...,

equals A000012 * A130123 * A000012, where A130123 = (1, 0,2; 0,0,4; 0,0,0,8; ...). Row sums of this triangle = A000337 starting (1, 5, 17, 49, 129, ...). (End)

First differences are A057728 = 1; 1; 1; 1; 2,1; 1; 4,2,1; 1; 8,4,2,1; 1; ... i.e., decreasing powers of 2, separated by another "1". - M. F. Hasler, May 06 2009

Apart from first term, numbers that are powers of 2 or the sum of some consecutive powers of 2. - Omar E. Pol, Feb 14 2013

From Andres Cicuttin, Apr 29 2016: (Start)

Numbers that can be digitally generated with twisted ring (Johnson) counters. This is, the binary digits of a(n) correspond to those stored in a shift register where the input bit of the first bit storage element is the inverted output of the last storage element. After starting with all 0’s, each new state is obtained by rotating the stored bits but inverting at each state transition the last bit that goes to the first position (see link).

Examples: for a(n) represented by three bits

Binary

a(5)= 4 -> 100 last bit = 0

a(6)= 6 -> 110 first bit = 1 (inverted last bit of previous number)

a(7)= 7 -> 111

and for a(n) represented by four bits

Binary

a(8) = 8 -> 1000

a(9) = 12 -> 1100 last bit = 0

a(10)= 14 -> 1110 first bit = 1 (inverted last bit of previous number)

a(11)= 15 -> 1111

(End)

Powers of 2 represented in bases which are terms of this sequence must always contain at least one digit which is also a power of 2. This is because 2^i mod (2^i - 2^j) = 2^j, which means the last digit always cycles through powers of 2 (or if i=j+1 then the first digit is a power of 2 and the rest are trailing zeros). The only known non-member of this sequence with this property is 5. - Ely Golden, Sep 05 2017

Numbers k such that k = 2^(1 + A000523(k)) - 2^A007814(k). - Daniel Starodubtsev, Aug 05 2021

A002260(n) = v(a(n)/2^v(a(n))+1) and A002024(n) = A002260(n) + v(a(n)) where v is the dyadic valuation (i.e., A007814). - Lorenzo Sauras Altuzarra, Feb 01 2023

The table is constructed so that row labels are 2^n - 1, and column labels are 2^n. The body of the table is the row*col + 1. A MAGMA program is provided that generates the numbers in a table format. The sequence is read along the antidiagonals starting from the top left corner.

All of these numbers have the following property:

let m be a member of A(n),

if a sequence B(n) = all i such that i XOR (m - 1) = i - (m - 1), then

the differences between successive members of B(n) is a repeating series

of 1's with the last difference in the pattern m. The number of ones in

the pattern is 2^j - 1, where j is the column index.

As an example consider A(4) which is 9,

the sequence B(n) where i XOR 8 = i - 8 starts as:

8, 9, 10, 11, 12, 13, 14, 15, 24... (A115419)

with successive differences of:

1, 1, 1, 1, 1, 1, 1, 9.

The main diagonal is the 6th cyclotomic polynomial evaluated at powers of two (A020515).

The formula for diagonals above the main diagonal

2^(2*n+1) - 2^(n + (a+1)/2) + 1 n>=(a+1)/2 a=odd number above diagonal

2^(2*n) - 2^(n + (b/2)) + 1 n>=(b/2)+1 b=even number above diagonal

The formulas for diagonals below the main diagonal

2^(2*n+1) - 2^(n + 1 -(a+1)/2) + 1 n>=(a+1)/2 a=odd number below diagonal

2^(2*n) - 2^(n - (b/2)) + 1 n>=(b/2)+1 b=even number below diagonal

Primes of this sequence are in A152449.

Conjecture: (i) Any integer n > 1 can be written as k + m with k > 0 and m > 0 such that 2^k*(2^{phi(m)} - 1) + 1 is prime.

(ii) Each integer n > 2 can be written as k + m with k > 0 and m > 0 such that 2^k*(2^{phi(m)} - 1) - 1 is prime.

Part (i) of the conjecture implies that this sequence has infinitely many terms. See also A234399.

Note that the sequence contains all Fermat primes and Mersenne primes since 2^k + 1 = 2^k*(2^{phi(1)} - 1) + 1 and 2^p - 1 = 2*(2^{phi(p)} - 1) + 1, where k is a positive integer and p is a prime.