cp's OEIS Frontend

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

If there are no odd perfect numbers, then the terms give all solutions n > 1 to A323244(n) = 0.

Conversely, if these are all numbers k > 1 that satisfy A323244(k) = 0 (which can be proved if one can show, for example, that no number in A007916 can satisfy the equation), then no odd perfect numbers exist. See also A336700. - Antti Karttunen, Jan 12 2024

Any Mersenne prime (A000668) times any power of 2, i.e., sequence A335431, is fixed by this map (note the indexing), including also all even perfect numbers. It is not currently known whether there are any additional fixed points.

Because a(n) has the same prime signature as A163511(n), it implies that applying A046523 and A052409 to this sequence gives the same results as with A163511, namely, sequences A278531 and A365805. - Antti Karttunen, Oct 09 2023

If there are no odd perfect numbers, then all n for which a(n) is 0 are given by sequence A324200.

Any Mersenne prime (A000668) times any power of 2 (i.e., 2^k, for k>=0) is fixed by this sequence, including also all even perfect numbers.

From Antti Karttunen, Jul 10 2020: (Start)

This is a "tuned variant" of A243071, and has many of the same properties.

For example, for n > 1, A007814(a(n)) = A007814(n) - A209229(n), that is, this map preserves the 2-adic valuation of n, except when n is a power of two, in which cases that value is decremented by one, and in particular, a(2^k * n) = 2^k * a(n) for all n > 1. Also, like A243071, this bijection maps primes to the terms of A000225 (binary repunits). However, the "tuning" (A332213) has a specific effect that each Mersenne prime (A000668) is mapped to itself. Therefore the terms of A335431 are fixed by this map. Furthermore, I conjecture that there are no other fixed points. For the starters, see the proof in A335879, which shows that at least none of the terms of A335882 are fixed.

(End)

Sequence A243071(A001599(k)), k >= 1, sorted into ascending order. See also comments in A324185.

Also positions of zeros in A324189.