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A348291
Number of pairs of positive numbers k and m < n such that n is the Nim-product of k and m.
1
0, 0, 1, 2, 1, 1, 2, 2, 5, 3, 5, 7, 10, 10, 12, 14, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 5, 5, 5, 5, 5, 3, 5, 3, 5, 5, 5, 5, 7, 7, 7, 7, 10, 10, 10, 10, 10, 10, 10, 10, 12, 12, 12, 12, 14, 14, 14, 14, 13, 13, 19, 21, 15
OFFSET
0,4
COMMENTS
This sequence reaches for all a(2^(2^n) - 1) a local maximum, because for all natural numbers n, the set of nimbers less than 2^(2^n) form the Galois field GF(2^(2^n)).
LINKS
Joseph DiMuro, On On_p, arXiv:1108.0962 [math.RA], 2015.
Wikipedia, Nimber
FORMULA
a(2^(2^n)..2^(2^n) + 2^(2^(n-1) - 1) - 1) = 1;
a(2^(2^n) - 1) = 2^(2^n) - 2;
EXAMPLE
Nim-product for 0..3:
0 1 2 3
-------
0|0 0 0 0
1|0 1 2 3
2|0 2 3 1
3|0 3 1 2 factors of 3 are 1 and 3.
PROG
(MATLAB)
function a = A348291( max_n )
a(1) = 0;
multable = zeros(max_n , max_n);
for n = 1:max_n
for m = 1:max_n
multable(m, n) = nimprod(m, n);
end
end
for n = 1:max_n
[i, j] = find(multable(1:n, 1:n) == n);
a(n+1) = length(find(unique([i j]) < n));
end
end
% highest power of 2 that divides a given number
function h2 = hpo2( in )
h2 = bitand(in, bitxor(in, 2^32-1)+1);
end
% base 2 logarithm of the highest power of 2 dividing a given number
function lhp2 = lhpo2( in )
lhp2 = 0;
m = hpo2(in);
q = 0;
while m > 1
m = m/2;
lhp2 = lhp2+1;
end
end
% nim-product of two numbers
function prod = nimprod(x, y)
if (x < 2 || y < 2)
prod = x * y;
else
h = hpo2(x);
if (x > h)
prod = bitxor(nimprod(h, y), nimprod(bitxor(x, h), y));
else
if (hpo2(y) < y)
prod = nimprod(y, x);
else
xp = lhpo2(x);
yp = lhpo2(y);
comp = bitand(xp, yp);
if (comp == 0)
prod = x * y;
else
h = hpo2(comp);
prod = nimprod(nimprod(bitshift(x, -h), bitshift(y, -h)), bitshift(3, (h - 1)));
end
end
end
end
end
CROSSREFS
Cf. A051775.
Sequence in context: A153902 A318205 A046772 * A359652 A179974 A246402
KEYWORD
nonn
AUTHOR
Thomas Scheuerle, Oct 10 2021
STATUS
approved