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%I #20 Mar 12 2022 13:34:56
%S 4,9,10,21,22,15,14,25,6,35,26,33,34,39,38,49,46,51,55,57,58,65,62,69,
%T 74,77,82,85,86,87,91,93,94,95,106,111,115,118,119,121,122,123,133,
%U 129,134,141,142,143,145,146,155,158,159,161,166,169,177,178,183,185,187
%N a(1)=4; thereafter a(n) is the smallest semiprime not yet appearing in the sequence that is relatively prime to a(n-1).
%C Every term is relatively prime to its neighbors.
%C Theorem: The sequence is a permutation of the semiprimes.
%C Proof (based on the arguments in The Yellowstone Permutation paper):
%C 1. The sequence is infinite. For P^2 is always a candidate for a(n), where P is any prime greater than all those dividing a(1),...,a(n-1).
%C 2. For any m, there is an n_0 such that a(n) > m for all n >= n_0. (Follows from 1.)
%C 3. For any prime p, there is a term divisible by p. Proof: Suppose p never divides any term of the sequence. Then no prime q>p can appear either, or else the first time there is a term q*r, we could have used p*r instead. So only finitely many primes appear, and so the sequence is finite, a contradiction.
%C 4. For any prime p, there are infinitely many terms divisible by p. Proof: Suppose there are only finitely many multiples of p, say p*q_1, p*q_2, ..., p*q_k. Let r,s,t be the next three primes after max{q_1,...,q_k}. None of p*r, p*s, p*t appear in the sequence. Choose n_0 so that a(n) > (p*t)^2 for n >= n_0. Suppose a(n) = x*y for some n > n_0. Then p*r, p*s, p*t are candidates for a(n+1) which are less than (p*t)^2, and since a(n) only involves two primes, one of the three is a smaller choice for a(n+1), a contradiction.
%C 5. For any prime p, there is a term a(n)=p^2. Proof: Similar to that of 4.
%C 6. Every semiprime appears. Proof: Let p*q be the smallest missing semiprime. Choose n_0 so that for n >= n_0, a(n) > (p*q)^2. Suppose a(n)=b*c with b <= c. Then a(n+1) will be p*q (and we have the desired contradiction) unless b is p or q. If b is p or q then a(n+2) = p*q unless a(n+2) is divisible by q or p, and so on. The only way that p*q will not appear is that for all n > n_0, a(n) is divisible alternately by p or q. But this contradicts 5, since there are infinitely many large prime squares in the sequence. QED. - _N. J. A. Sloane_, Oct 13 2015
%H David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, <a href="http://arxiv.org/abs/1501.01669">The Yellowstone Permutation</a>, arXiv preprint arXiv:1501.01669, 2015. Also Journal of Integer Sequences, Vol. 18 (2015), Article 15.6.7
%t sp0=Select[Range[1000],2==Plus@@Last/@FactorInteger@#&]; sp=sp0; le=Length@sp; seq={4}; b=4; sp=Rest@sp; le=le-1; Do[Do[spi=sp[[i]]; If[1==GCD[b,spi],b=spi; AppendTo[seq,b]; sp=Delete[sp,i]; le=le-1; Break[]],{i,le}],{100}]; seq
%Y Cf. A098550.
%Y A001358 gives the semiprimes.
%K nonn
%O 1,1
%A _Zak Seidov_, Jun 13 2006
%E Definition revised by _N. J. A. Sloane_, Oct 13 2015 at the suggestion of _Bob Selcoe_.
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