[math-fun] First consecutive run of composites of a given length
A well-known guaranteed way to get a run of N consecutive composite integers is to multiply together all primes <= N+1, and then add 2, 3, 4, . . ., N+1, respectively, to the product. E.g., to get 9 consecutive composites, we could use [210+2 . . 210+10] = [212 . . 220] (since 210 = 2*3*5*7 with 7 <= 9+1 < 11). Call this sequence S(N). Which suggests a few questions: 1. How often, does S(N) give the *first* run of N composites? ([114 . . 122] is the first run of 9, rather than S(9) = [212 . . 220].) 2. How much farther can S(N) be extended to include more consecutive composites? (S(9) extends 2 farther to [212 . . 222].) And especially 3. {4 (1), 8 (3), 24 (5), 30 (7), 114 (13), . . . } shows the first number of each new record-breaking (in length) run of composites followed by the (length) of this run. The latter sequence 1, 3, 5, 7, 13,... is A066963 in OEIS, but 4, 8, 24, 30, 114, . . . -- the first numbers of each such run -- does not seem to be there. So I wonder if there is an asymptotic expression for A066963, or the companion sequence 4, 8, 24, 30, 114, . . . (or more likely their logs). (Or maybe a "fairer" companion sequence should be the *centers* of each record run of composites: 4, 9, 26, 33, 120, . . . ?) --Dan
Note that a run of (exactly/at least) n composites corresponds to a prime gap of (exactly/at least) n+1. Most of the work in this area has focussed on the primes, not the composites. ________________________________________________________________________ Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading spam and email virus protection.
participants (2)
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Daniel Asimov -
franktaw@netscape.net