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A213253
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a(n) = smallest k such that highest prime factor of m(m+1)...(m+k-1) is > n if m > n.
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2
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1, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14
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OFFSET
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1,2
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COMMENTS
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By a theorem of Sylvester, a(n) always exists.
Najman says that standard heuristics for the size of gaps between consecutive primes lead one to expect that the order of magnitude of a(n) is (log n)^2. - Jonathan Sondow, Jul 23 2013
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REFERENCES
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E. F. Ecklund, Jr., R. B. Eggleton and J. L. Selfridge, Factors of consecutive integers, Proc. Man. Conference Numerical Maths., Winnipeg, (1971), 155-157.
E. F. Ecklund, Jr., R. B. Eggleton and J. L. Selfridge, Consecutive integers all of whose prime factors belong to a given set, Proc. Man. Conference Numerical Maths., Winnipeg (1971), 161-162.
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LINKS
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FORMULA
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a(n) <= n (Sylvester's theorem--see Sylvester 1892, p. 4) - Jonathan Sondow, Jul 23 2013
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MATHEMATICA
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(* To speed up computation, it is assumed that a(n) >= a(n-1)-2 and m <= n^2 *) a[1] = 1; a[n_] := a[n] = For[k = a[n-1]-2, True, k++, If[And @@ (FactorInteger[ Pochhammer[#, k]][[-1, 1]] > n & /@ Range[n+1, n^2]), Return[k]]]; Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 1, 268}] (* Jean-François Alcover, Nov 25 2013 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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