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 A194585 Starting points of stapled intervals of length 17. 3
 2184, 27830, 32214, 57860, 62244, 87890, 92274, 117920, 122304, 147950, 152334, 177980, 182364, 208010, 212394, 238040, 242424, 268070, 272454, 298100, 302484, 328130, 332514, 358160, 362544, 388190, 392574, 418220, 422604, 448250 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS "Stapled" intervals are defined in A090318. They are at least of length 17, and those of this minimal length are listed here. Therefore, this is not only a subsequence of A130173, but also of A130171. From Fidel I. Schaposnik, Aug 16 2014: (Start) Let S be the set of distinct prime factors appearing in the factorization of at least two different numbers in the range [a,b], and m the product of all the elements in S. Then it is clear that if [a,b] is a stapled interval, so is [m+a,m+b]. Moreover, if a > m then the range [a-m,b-m] is also a stapled interval of the same length, so we can group the stapled intervals of a given length in "chains". To prove the g.f., note that S cannot contain any prime number greater or equal to b-a+1, so for stapled intervals of length 17 the maximum value of m is m = 2*3*5*7*11*13 = 30030. Then any stapled interval of length 17 must belong to a chain whose first element is at most 30030, and the only stapled intervals in this range are [2184,2200] and [27830,27846]. The g.f. encompasses both these chains, namely a(2*n+1) = 2184 + 30030*n and a(2*n+2) = 27830 + 30030*n. (End) LINKS Fidel I. Schaposnik, Table of n, a(n) for n = 1..666 FORMULA From Colin Barker, Aug 16 2014: (Start) a(n) = (-15031+10631*(-1)^n+30030*n)/2. a(n) = a(n-1)+a(n-2)-a(n-3). G.f.: 2*x*(1100*x^2+12823*x+1092) / ((x-1)^2*(x+1)). (End) PROG (PARI) {u=vector(17, j, 1); v=vector(17, j, j); for(k=2, 1e9, nextprime(k)

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Last modified July 11 04:33 EDT 2020. Contains 335609 sequences. (Running on oeis4.)