A225822 Lesser of adjacent odd numbers with different parity of binary weight and both isolated from odd numbers of same parity of binary weight.

7, 23, 31, 39, 55, 71, 87, 95, 103, 119, 127, 135, 151, 159, 167, 183, 199, 215, 223, 231, 247, 263, 279, 287, 295, 311, 327, 343, 351, 359, 375, 383, 391, 407, 415, 423, 439, 455, 471, 479, 487, 503, 511, 519, 535, 543, 551, 567, 583, 599, 607, 615, 631

Offset: 1

Author

Brad Clardy, Jul 30 2013

Keywords

• nonn
• base
• easy

Comments

Write the sequence of odious odd numbers above the sequence of evil odd numbers connecting all that are 2 apart:

1 7 11-13 19-21 25 31 35-37 41 47-49 55 59-61 67-69 73 79-81 87 91-93 97

3-5 9 15-17 23 27-29 33 39 43-45 51-53 57 63-65 71 75-77 83-85 89 95 99-

Remove the connected numbers:

1 7 25 31 41 55 73 87 97

9 23 33 39 57 71 89 95

Define these as "isolated".

The sequence is the smaller of the remaining pairs that are 2 apart.

The 1 is not a member since there is no change in parity between 1 and 7.

All of the differences between adjacent numbers in both the evil and odious sequences are either 2, 4 or 6, with 4 being the indicator that a transition in parity occurs. The program provided utilizes that fact to produce the sequence.

The sequence that includes all numbers along this path is A047522 (numbers congruent to {1,7} mod 8). This is also the same as the odd terms of A199398 (XOR of the first n odd numbers).

This sequence is similar to A044449 (numbers n such that string 1,3 occurs in the base 4 representation of n but not of n+1), but it contains additional terms. An example is 119. Its base 4 representation is 1313 while the base 4 representation of 120 is 1320. It may be that another workable definition of the sequence is -- numbers n in base 4 representation such that string 1,3 occurs one less time in n+1 than n, but I have not been able to check this.

The difference between the numbers in the sequence is always either 8 or 16, however there appears to be no recurring repetitions in it. Writing the 8 as a 0 and the 16 as a 1 (or dividing the difference pattern by 2 and subtracting a 1) produces a difference pattern of: 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1... which is an infinite word.

A similar process writing Even Odious over Even Evils produces 6, 22, 30, 38, 54, 70... which is twice A131323 (Odd numbers n such that the binary expansion ends in an even number of 1's), with all numbers along the path given by A047451 (numbers congruent to {0,6} mod 8) and yields the same difference pattern which produces the same infinite word.

Links

• Brad Clardy, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A001969 (evil numbers), A129771 (odd evil numbers).

Cf. A000069 (odious numbers), A092246 (odd odious numbers).

Cf. A047522 (numbers congruent to {1,7} mod 8).

Cf. A199398 (XOR of first n odd numbers).

Cf. A044449 (a subset of this sequence).

Cf. A131323 (odd numbers n such that the binary expansion ends in an even number of 1's).

Cf. A047451 (numbers congruent to {0,6} mod 8).

Cf. A000120 (binary weight of n).

Cf. A079523.

Programs

Magma

//Function Bweight calculates the binary weight of an integer
Bweight := function(m)
Bweight:=0;
adigs := Intseq(m,2);
for n:= 1 to Ilog2(m)+1 do
  Bweight:=Bweight+adigs[n];
end for;
return Bweight;
end function;
prevodi:=0;
currodi:=0;
m:=0;
count:=0;
for n:= 1 to 20000 by 2 do
    m:=m+1;
    if (Bweight(n) mod 2 eq 1) then odious:=true; else odious:=false; end if;
    if (odious) then currodi:=n; end if;
       if (currodi - prevodi eq 4) then
          if (m mod 2 eq 1) then count:=count+1; count,n-2;
             else count:=count+1;count,n-4;
       end if;
    end if;
    if(odious) then prevodi:=currodi; end if;
end for;

Mathematica

2*Select[Range[1, 320, 2], EvenQ[IntegerExponent[# + 1, 2]] &] + 1 (* Amiram Eldar, Jul 24 2023 *)

PARI

is(n)=n%4==3 && valuation(n\4+1, 2)%2 \\ Charles R Greathouse IV, Aug 20 2013

Python

from itertools import count, islice
def A225822_gen(startvalue=1): # generator of terms >= startvalue
    return map(lambda m:(m<<1)+1,filter(lambda n:n&1 and not (~(n+1)&n).bit_length()&1,count(max(startvalue,1))))
A225822__list = list(islice(A225822_gen(),30)) # Chai Wah Wu, Jul 09 2022

Formula

a(n) = 2*A131323(n) + 1.

a(n) = 4*A079523(n) + 3. - Charles R Greathouse IV, Aug 20 2013

a(n) ~ 12n. (In particular, a(n) = 12n + O(log n).) - Charles R Greathouse IV, Aug 20 2013