Thomas Scheuerle, <a href="/A348369/a348369.svg">a(1)..a(4000)</a> (Both axes are logarithmic and denote 2^x and 2^y. It appears that this sequence is self-similar , with an irrational exponent.)

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Thomas Scheuerle, <a href="/A348369/a348369.svg">a(1)..a(4000)</a> (Both axes are logarithmic and denote 2^x and 2^y. It appears that this sequence is self-similar with an irrational exponent).)

proposed

editing

editing

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Thomas Scheuerle, <a href="/A348369/a348369.svg">a(1)..a(4000) </a> (Both axes are logarithmic and denote 2^x and 2^y. It appears that this sequence is self-similar with an irrational exponent).</a>

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editing

editing

proposed

Number of ways A328596(n) (the reversed binary expansion is an aperiodic necklace) can be expressed as sum A328596(k) + A328596(m) with 0 < k,m < n. The cases A328596(k) + A328596(m) and A328596(m) + A328596(k) are considered as equal.

Conjecture: The only zero in this sequence is a(1). A348268 maps all elements terms of A328596 bijective to primes. Let P be this bijection between Lyndon words and primes and P' its inverse. Then for each prime q, there exist primes r and s such that q = P(P'(r) + P'(s)). If we would were to define a table T(m,n) which encodes the sum q + 1 = (A000040(m) + A000040(n)), then q = P(P'(A000040(m)) + P'(A000040(n))) would be a permutation of this table, ; this connects this conjecture to Goldbach's conjecture.

All reversed binary expansions of powers of two are Lyndon words. All reversed binary expansions of numbers of the form 2*(2^m - 1) are Lyndon words , too. 2*(2^m - 1) + 2 is again a power of 2. All natural numbers Every positive integer can be expressed as a sum of powers of 2. From this we can conclude, that it is always possible to compose elements terms of A328596(n) (n > 1), as a sum of elements terms of A328596. This would require at least 2 or more such elementsterms.

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All reversed binary expansions of powers of two are Lyndon words. All reversed binary expansions of numbers of the form 2*(2^m - 1) are Lyndon words too. 2*(2^m - 1) + 2 is again a power of two2. All natural numbers can be expressed as a sum of powers of two2. From this we can conclude, that it is always possible to compose elements of A328596(n) (n > 1), as a sum of elements of A328596. This would require at least two 2 or more such elements.

Tabel1Table A: A348268(A348268^-1(m) + A348268^-1(n))

Tabel2Table B: m + n

Tabel2 Table B is a permutation of Tabel1 Table A + 1.

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