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 than 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
Index entries for linear recurrences with constant coefficients, signature (1, 1, -1).
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)<k+17 & (v+=u*(-v[1]+k=precprime(k+17))) & next; v+=u; for(j=k, k+16, vecsort(gcd(j, v), , 8)[2]<j | next(2)); print1(k", "))}
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Oct 14 2011
STATUS
approved