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 A164799 a(n) = the smallest positive integer such that the product of a(n) consecutive integers, where n is the smallest, is divisible by every prime from 2 to the largest prime divisor of the product. (a(1)=1.) 5
 1, 1, 2, 1, 2, 1, 4, 1, 2, 5, 5, 1, 10, 2, 2, 1, 10, 1, 16, 2, 18, 17, 16, 1, 22, 21, 20, 19, 18, 1, 28, 1, 30, 29, 2, 1, 26, 37, 36, 35, 34, 41, 40, 43, 42, 41, 40, 1, 46, 45, 44, 43, 42, 1, 52, 51, 50, 49, 48, 1, 58, 61, 4, 1, 58, 57, 56, 67, 66, 65, 64, 1, 70, 73, 72, 71, 70, 69, 68, 2 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS a(n) = A164798(n) - n +1. LINKS Table of n, a(n) for n=1..80. EXAMPLE Consider the products of consecutive integers, (m+9)!/9!, m >= 1. First, 10 is divisible by 2 and 5, but there is a prime gap since 3 is missing from the factorization. 10*11 is divisible by 2, 5, and 11, but 3 and 7 are missing. 10*11*12 is divisible by 2, 3, 5, and 11, but 7 is missing. 10*11*12*13 is divisible by all primes up to 13, except 7. But 10*11*12*13*14 is indeed divisible by every prime from 2 to 13. So a(10) = 5 because 5 consecutive numbers are multiplied together. MAPLE Contribution from R. J. Mathar, Feb 27 2010: (Start) A000040v := proc(pmax) L := {} ; for i from 1 do if ithprime(i) <= pmax then L := L union {ithprime(i)} ; else return L; end if end do ; end: A164799 := proc(n) local k, p ; if n = 1 then return 1 ; end if; for k from 1 do p := ifactors(mul(n+i, i=0..k-1))[2] ; p := {seq(op(1, d), d=p)} ; pL := A000040v(max(op(p))) ; if p = pL then return k; end if; end do ; end proc: seq(A164799(n), n=1..90) ; (End) CROSSREFS A164798 Sequence in context: A256607 A256608 A279186 * A274451 A193787 A072614 Adjacent sequences: A164796 A164797 A164798 * A164800 A164801 A164802 KEYWORD nonn AUTHOR Leroy Quet, Aug 26 2009 EXTENSIONS Terms beyond a(13) from R. J. Mathar, Feb 27 2010 STATUS approved

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Last modified May 20 06:19 EDT 2024. Contains 372703 sequences. (Running on oeis4.)