A358919 -id:A358919 - cp's OEIS frontend

A358918 a(0) = 0, and for any n >= 0, a(n+1) is the length of the longest run of consecutive terms a(i), ..., a(j) with 0 <= i <= j <= n such that a(i) XOR ... a(j) = a(n) (where XOR denotes the bitwise XOR operator).

0, 1, 2, 1, 2, 4, 6, 2, 7, 9, 1, 6, 7, 9, 12, 9, 12, 13, 14, 17, 1, 14, 17, 18, 20, 18, 20, 22, 26, 18, 20, 28, 30, 26, 22, 28, 30, 30, 32, 3, 28, 30, 32, 3, 28, 38, 40, 22, 34, 46, 26, 41, 31, 45, 42, 40, 37, 41, 31, 58, 60, 53, 54, 57, 57, 57, 57, 57, 57, 57

Offset: 0

Rémy Sigrist, Dec 06 2022

The sequence has periodic behavior over large intervals (see illustrations in Links section).
This sequence is unbounded:
- let b(k) = a(0) XOR ... XOR a(k),
- if the sequence was bounded with greatest value m,
- then the sequence b would be bounded by m*(m+1)/2,
- and some value in b, say v, would appear infinitely many times,
- so we can find occurrences of v in b at distance m' > m,
  say b(k) = b(k + m') = v,
- so a(k+1) XOR a(k+2) XOR ... XOR a(k+m') = 0,
- and a(k+1) XOR a(k+2) XOR ... XOR a(k+m'+1) = a(k+m'+1),
  and a(k+m'+2) >= m' > m, a contradiction.

			The first terms, alongside an appropriate pair (i,j), are:
  n   a(n)  (i,j)
  --  ----  -------
   0     0  N/A
   1     1  (0,0)
   2     2  (0,1)
   3     1  (2,2)
   4     2  (0,1)
   5     4  (0,3)
   6     6  (0,5)
   7     2  (4,5)
   8     7  (0,6)
   9     9  (0,8)
  10     1  (9,9)
  11     6  (2,7)
  12     7  (2,8)

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Thomas Scheuerle, log_2(abs(first differences of a(0...2000)))
Thomas Scheuerle, a(0...2000) colored by the sign of the first differences
Rémy Sigrist, C program
Rémy Sigrist, Scatterplot of the first 1000000 terms (numbers correspond to "local" periods)

Cf. A358799, A358919.

C
```
See Links section.
```

function a = A358918( max_n )
    a = 0; xr = 0;
    for n = 2:max_n
        r = 0;
        xr(n) = bitxor(xr(n-1),a(end));
        f = find(xr==a(end),1,'last');
        if ~isempty(f)
            r = f-1;
        end
        x = xr(2:end);
        for k = length(a)-1:-1:1
            x(1:k) = bitxor(x(1:k),ones(1,k)*a(length(a)-k));
            x(1:k) = bitxor(x(1:k),a([1:k]+(length(a)-k)));
            f = find(x==a(end),1,'last');
            if ~isempty(f)
                if r < f
                    r = f;
                end
            end
        end
        a(n) = r;
    end
end % Thomas Scheuerle, Dec 08 2022

A359034 a(n+1) is the sum of the number of terms in all groups of contiguous terms that add up to a(n); a(1)=1.

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 4, 5, 3, 5, 4, 6, 6, 7, 7, 8, 10, 11, 4, 7, 9, 9, 10, 12, 13, 14, 13, 15, 8, 11, 7, 10, 13, 16, 19, 18, 18, 19, 19, 20, 7, 11, 8, 12, 14, 14, 15, 9, 11, 9, 12, 15, 10, 14, 16, 20, 14, 17, 17, 18, 22, 22, 23, 22, 24, 23, 23, 24, 24, 25, 28, 27, 22

Offset: 1

Neal Gersh Tolunsky, Dec 12 2022

nonn

If strongly smoothened, this sequence displays growth. This growth appears to be caused by the number of groups which is increasing by growing length of the sequence roughly proportional to n^(1/2). But the length of the groups appears to be nearly uninfluenced by this. - Thomas Scheuerle, Dec 14 2022

			a(17) is 10 because in the sequence so far (1, 1, 2, 3, 3, 4, 4, 5, 3, 5, 4, 6, 6, 7, 7, 8), these are the ways of adding contiguous terms to get a(16)=8: (2, 3, 3); (4, 4); (5, 3); (3, 5); (8). This is 10 terms in total, so a(17) is 10. Notice groups (5,3) and (3,5) overlap.

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Thomas Scheuerle, Scatter plot of a(n) per number of groups found. For n = 1...20000. This is the mean number of contiguous terms for each n.
Thomas Scheuerle, Scatter plot, number of groups of contiguous terms for n = 1...20000

Cf. A331614, A358919. Begins the same as A124056 (until a(13)).

function a = A359034( max_n )
    a = [1 1];
    for n = 3:max_n
        s = 1; e = 1; sm = 1; c = 0;
        while e < n-1
            while sm < a(n - 1) && e < (n - 1)
                e = e + 1; sm = sm + a(e);
            end
            if sm == a(n - 1)
                c = c + (e - s) + 1;
            end
            s = s + 1;
            e = s; sm = a(s);
        end
        a(n) = c + 1;
    end
end % Thomas Scheuerle, Dec 14 2022

cp's OEIS Frontend

A358918 a(0) = 0, and for any n >= 0, a(n+1) is the length of the longest run of consecutive terms a(i), ..., a(j) with 0 <= i <= j <= n such that a(i) XOR ... a(j) = a(n) (where XOR denotes the bitwise XOR operator).

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