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approved
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Revision History for A059894
(Bold, blue-underlined text is an addition; faded, red-underlined text is aShowing entries 1-10 | older changes
Complement and reverse the order of all but the most significant bit in binary expansion of n. n = 1ab..yz -> 1ZY..BA = a(n), where A = 1-a, B = 1-b, ... .
(history; published version)
(history; published version)
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proposed
reviewed
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editing
proposed
FORMULA
To verify first conjecture above we use n = Sum_{k=0..l(n)} T(n,k)*2^k where l(n) = floor(log_2(n)), T(n,k) = floor(n/2^k) mod 2, so by definition we have a(n) = Sum_{k=0..l(n)} (1-T(n,k-1))*2^(l(n)-k), then we use l(n) = l(floor(n/2)) + 1, T(2n,k) = T(n,k-1), T(2n+1,k) = T(n,k-1) + [k=0] and apply shifting of the summation. - Mikhail Kurkov, Nov 11 2019 [verification needed]
STATUS
approved
editing
STATUS
proposed
approved
STATUS
editing
proposed
FORMULA
To verify first conjecture above we use n = Sum_{k=0..l(n)} T(n,k)*2^k where l(n) = floor(log_2(n)), T(n,k) = floor(n/2^k) mod 2, so by definition we have a(n) = Sum_{k=0..l(n)} (1-T(n,k-1))*2^(l(n)-k), then we use l(n) = l(floor(n/2)) + 1, T(2n,k) = T(n,k-1), T(2n+1,k) = T(n,k-1) + [k=0] and apply shifting of the summation. - Mikhail Kurkov, Nov 11 2019 [verification needed]
STATUS
approved
editing
STATUS
editing
approved
LINKS
Harry J. Smith and Alois P. Heinz, <a href="/A059894/b059894.txt">Table of n, a(n) for n = 1..8191</a> (first 1024 terms from Harry J. Smith)
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approved
editing
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reviewed
approved