cp's OEIS Frontend

Prime factorization sentinel computed for A277699. Factorization is essentially done in the polynomial ring GF(2)[X].

A277807 lists the "sporadic" solutions to A048720(n, A065621(n)) = k^2, i.e., those solutions where n is not in A023758 (not yet proved: these are precisely those solutions where k <> n). This sequence gives the corresponding values of k.

If n is included in this sequence, then also 2n is included (and vice versa), thus the sequence is infinite and wholly determined by its odd terms.

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

a(n) = the least number with the same prime signature as A091203(n).

This sequence works as an A046523-analog in the polynomial ring GF(2)[X] and can be used as a filter which matches with (and thus detects) any sequence in the database where a(n) depends only on the exponents of irreducible factors when the polynomial corresponding to n (via base-2 encoding) is factored over GF(2). These sequences are listed in the Crossrefs section, "Sequences that partition N into ...".

Matching in this context means that the sequence a matches with the sequence b iff for all i, j: a(i) = a(j) => b(i) = b(j). In other words, iff the sequence b partitions the natural numbers to the same or coarser equivalence classes (as/than the sequence a) by the distinct values it obtains.

Not yet proved: Equally, numbers n such that A048720(n, A065621(n)) = k^2 for some k different from n.

If n is included in this sequence, then also 2n is included (and vice versa), thus the sequence is infinite and wholly determined by its odd terms.