OFFSET
0,3
COMMENTS
This sequence is conjectured to be a permutation of the nonnegative integers.
The second row of the XOR-triangle ((a(0) XOR a(1)), (a(1) XOR a(2)), ...) is a permutation of the positive integers. Further rows miss additional numbers. The first diagonal of this triangle, 0, 1, 2, 7, 3, 5, ..., (A345237) is not a permutation, because it contains multiple equal values.
If we enumerate the imaginary units of a Cayley-Dickson algebra of order m by the positive integers 1 .. 2^m-1, XOR of any pair of these numbers represents the multiplication table of this algebra if signs are ignored. This explains why a(2^n) and A345237(2^n) are equal.
The antidiagonals starting at a(0..2) are {0}, {1, 1}, {2, 3, 2}. Let D(n) be the antidiagonal starting at a(n). Then XOR-sum(D(2^n-1)) = 0 and XOR-sum(D((2*m + 1)*2^(n+1)-1)) = XOR-sum(D((2*m + 1)*2^(p+1)-1)). XOR-sum means XOR over all elements. This property holds also for the XOR-triangle based on the nonnegative integers ordered in sequence.
LINKS
Rémy Sigrist, Table of n, a(n) for n = 0..7448
Thomas Scheuerle, Triangle based on n = 0..149, drawn as colored surfaces bitwise for bit 0-7. This shows interesting Sierpinski-like structures.
Thomas Scheuerle, Triangle based on n = 0..149, colored by number.
Thomas Scheuerle, Triangle based on n = 0..149, colored by parity. Interestingly the amounts of red and green are approximately equal.
Rémy Sigrist, C++ program.
FORMULA
a(2^n) = A345237(2^n).
a(2^m + 2^n + 2^p + ...) = A345237(k) XOR A345237(k - 2^m) XOR A345237(k - 2^n) XOR A345237(k - 2^p) XOR A345237(k - 2^m - 2^n) XOR A345237(k - 2^m - 2^p) XOR A345237(k - 2^m - 2^n - 2^p) XOR ..., k = 2^m + 2^n + 2^p + ... .
Sum_{k=0..n} a(k) <= Sum_{k=0..n} A345237(k).
( Sum_{k=0..n} a(k) + Sum_{k=0..n} A345237(k) )^0.4202... < n and > n - 30 at least for n < 500.
EXAMPLE
0 1 2 4 3 7 5 8 16 ... <-- sequence
\ / \ / \ / \ / \ / \ / \ / \ / a(0),a(1),a(2),...
1 3 6 7 4 2 13 24 ... <----------+
\ / \ / \ / \ / \ / \ / \ / |
2 5 1 3 6 15 21 ... <-+ 2nd row is
\ / \ / \ / \ / \ / \ / | a(0) XOR a(1),
7 4 2 5 9 26 ... | a(1) XOR a(2),
\ / \ / \ / \ / \ / | a(2) XOR a(3),
3 6 7 12 19 ... | etc.
\ / \ / \ / \ / |
5 1 11 31 ... |
\ / \ / \ / |
4 10 20 ... |
\ / \ / |
14 30 ... 3rd row is
\ / (a(0) XOR a(1)) XOR (a(1) XOR a(2)),
16 ... (a(1) XOR a(2)) XOR (a(2) XOR a(3)),
etc.
We show why a(2^n) = A345237(2^n) by reproducing the same process with a randomly chosen set of octonion units: {e0,e1,e2,e5,e6}. XOR is replaced by multiplication.
e0 e1 e2 e5 e6
\/ \/ \/ \/
e1 e3 e7 -e3
\/ \/ \/
-e2 -e4 -e4
\/ \/
-e6 -e0
\/
e6
PROG
(MATLAB)
function a = A346298(max_n)
a(1) = 0;
for n = 1:max_n
t = 1;
while ~isok([a t])
t = t+1;
end
a = [a t];
end
end
function [ ok ] = isok( in )
ok = (length(in) == length(unique(in)));
x = in;
if ok
for k = 1:(length(x)-1)
x = bitxor(x(1:end-1), x(2:end));
ok = ok && (length(x) == length(unique(x)));
if ~ok
break;
end
end
end
end
(C++) See Links section.
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Thomas Scheuerle, Jul 13 2021
STATUS
approved