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A357574
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a(n) is the maximum number of pairs that sum to a power of 2 in a set of n consecutive odd numbers.
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2
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0, 1, 2, 4, 5, 7, 9, 11, 13, 15, 17, 19, 21, 24, 26, 29, 31, 34, 36, 39, 41, 44, 46, 49, 51, 54, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 150, 153, 157, 160, 164, 167
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OFFSET
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1,3
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COMMENTS
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An optimal set delivering a(n) pairs summing to powers of 2 can be formed by n-A357409(n) first negative odd numbers and A357409(n) first positive odd numbers, that is, by the odd numbers in the interval [-2*(n-A357409(n))+1, 2*A357409(n)-1].
a(n) is in many cases equal to A347301(n) but there are some deviations: a(3) = 2 but A347301(3) = 3, a(29) = 62 but A347301(29) = 61, a(31) = 68 but A347301(32) = 67, a(33) = 74 but A347301(33) = 73, ... . Hence it appears that a(n) may be used as an improved lower bound for A352178(n) in many cases.
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LINKS
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FORMULA
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a(n) >= n-1. This would be the maximum value that could be attained for a set of only positive odd numbers and size n.
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EXAMPLE
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a(5) = 5 because A357409(5) = 4, for which the corresponding set {-1, 1, 3, 5, 7} produces 5 powers of 2: 1+3, 1+7, 3+5, 3-1, 5-1.
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PROG
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(MATLAB)
a(1) = 0; q = [];
for n = 1:max_n
c = 0;
for k = 0:n
s = (2*([0:n]-k))+1;
r = countpowtwo(s);
if c < r
c = r;
q = s;
end
end
a(n+1) = c;
end
end
function c = countpowtwo(s)
M = repmat(s, [length(s), 1]);
M = M+M';
M(M<=0) = 7;
M = bitand(M, M-1);
M = M + eye(size(M));
c = length(find(M == 0))/2;
end
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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