OFFSET
1,5
COMMENTS
Conjecture: The only zero in this sequence is a(1). A348268 maps all 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 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. 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 terms of A328596(n) (n > 1), as a sum of terms of A328596. This would require at least 2 or more such terms.
LINKS
Thomas Scheuerle, a(1)..a(4000) (Both axes are logarithmic and denote 2^x and 2^y. It appears that this sequence is self-similar, with an irrational exponent.)
EXAMPLE
.
2 3 5 7
-----------------
2| (3) 4 6 8 prime numbers are marked by ()
3| 4 (5) (7)(11)
5| 6 (7)(11) 9
7| 8 (11) 9 (13)
.
Table B: m + n
2 3 5 7
-----------------
2| (4) 5 7 9 prime numbers + 1 are marked by ()
3| 5 (6) (8) 10
5| 7 (8) 10 (12)
7| 9 10 (12)(14)
.
Table B is a permutation of Table A + 1.
CROSSREFS
KEYWORD
nonn
AUTHOR
Thomas Scheuerle, Oct 15 2021
STATUS
approved