This site is supported by donations to The OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A119718 a(1)=4; thereafter a(n) =  smallest semiprime not yet appearing in the sequence that is relatively prime to a(n-1). 2
 4, 9, 10, 21, 22, 15, 14, 25, 6, 35, 26, 33, 34, 39, 38, 49, 46, 51, 55, 57, 58, 65, 62, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95, 106, 111, 115, 118, 119, 121, 122, 123, 133, 129, 134, 141, 142, 143, 145, 146, 155, 158, 159, 161, 166, 169, 177, 178, 183, 185, 187 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Every term is relatively prime to its neighbors. Theorem: The sequence is a permutation of the semiprimes. Proof (based on the arguments in The Yellowstone Permutation paper): 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). 2. For any m, there is an n_0 such that a(n) > m for all n >= n_0. (Follows from 1.) 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. 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. 5. For any prime p, there is a term a(n)=p^2. Proof: Similar to that of 4. 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 LINKS David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, The Yellowstone Permutation, arXiv preprint arXiv:1501.01669, 2015. Also Journal of Integer Sequences, Vol. 18 (2015), Article 15.6.7 MATHEMATICA 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 CROSSREFS Cf. A098550. A001358 gives the semiprimes. Sequence in context: A004631 A316113 A245096 * A263648 A051884 A131368 Adjacent sequences:  A119715 A119716 A119717 * A119719 A119720 A119721 KEYWORD nonn AUTHOR Zak Seidov, Jun 13 2006 EXTENSIONS Definition revised by N. J. A. Sloane, Oct 13 2015 at the suggestion of Bob Selcoe. STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified September 21 11:45 EDT 2019. Contains 327253 sequences. (Running on oeis4.)