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A293975
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If n is even, divide by 2; otherwise, add the next larger prime.
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4
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0, 3, 1, 8, 2, 12, 3, 18, 4, 20, 5, 24, 6, 30, 7, 32, 8, 36, 9, 42, 10, 44, 11, 52, 12, 54, 13, 56, 14, 60, 15, 68, 16, 70, 17, 72, 18, 78, 19, 80, 20, 84, 21, 90, 22, 92, 23, 100, 24, 102, 25, 104, 26, 112, 27, 114, 28, 116, 29, 120, 30, 128, 31, 130, 32, 132, 33, 138, 34
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OFFSET
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0,2
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COMMENTS
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Inspired by the "PrimeLatz" map A174221 (where the next three primes are added).
The trajectory under iterations of this map seems to end in the cycle 1 -> 3 -> 8 -> 4 -> 2 -> 1, for any starting value n. Can this be proved?
In order to develop a proof, one can consider the "condensed" version of the map which is: h(x) = odd_part(x+nextprime(x)); i.e., add the next prime, then remove all factors of 2. It is easy to see that this map verifies, for all x > 2, h(x) <= x + g(x)/2 where g(x) is the gap between the x and the next larger prime. Often, h(x) will be close to x/2 or even to x/4 or smaller. Nonetheless, for any power (iteration) of h, there are numbers for which h^m is increasing, e.g., h(h(h(x))) > x for x = 1, 525, 891, 1071, 1135, ..., and h^4(x) > x for x = 2, 1329, 5591, 8469, 9555, ...
It suffices to prove that if n > 1 is odd, the trajectory {x(i)} starting at x(0)=n contains some number < n. Let p = nextprime(n). As long as x(2k) is odd we have x(2k+1) = x(2k)+p and x(2k+2)=(x(2k)+p)/2 with
n <= x(2k) < x(2k+2) < p. But this can only continue finitely many times: eventually x(2k) must be even, and then x(2k+1) < p/2 < n (by Bertrand's postulate). (End)
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LINKS
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MAPLE
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seq(op([k, 2*k+1+nextprime(2*k+1)]), k=0..100); # Robert Israel, Nov 08 2017
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MATHEMATICA
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Array[If[EvenQ@ #, #/2, NextPrime@ # + # &@ #] &, 69, 0] (* Michael De Vlieger, Nov 08 2017 *)
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PROG
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(PARI) A293975(n)=if(bittest(n, 0), n+nextprime(n+1), n\2)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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