

A049300


Smallest number starting a longest interval of consecutive integers, each of which is divisible by at least one of the first n primes.


3



2, 2, 2, 2, 114, 9440, 217128, 60044, 20332472, 417086648, 74959204292, 187219155594, 79622514581574, 14478292443584, 6002108856728918, 12288083384384462, 5814429911995661690, 14719192159220252523420
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OFFSET

1,1


COMMENTS

The length of such longest interval of consecutive integers is given by A058989(n), which is the first maximal gaps A048670(n) minus 1 in the reduced residue system of consecutive primorial numbers.
Let j(m)=A048669(m) be the Jacobsthal function, i.e., the maximal distance between integers relatively prime to m. Let m=2*3*5*...*prime(n). Then a(n) is the least k>0 such that k,k+1,k+2,...k+j(m)2 are not coprime to m. Note that a(n) begins (or is inside) a large gap between primes.  T. D. Noe, Mar 29 2007


LINKS

Max Alekseyev, Table of n, a(n) for n = 1..24


FORMULA

a(n) = 1 + A128707(A002110(n)).  T. D. Noe, Mar 29 2007


EXAMPLE

Between 1 and 7, all 5 numbers (2,3,4,5,6) are divisible either by 2,3 or 5. Thus a(3)=2, the initial term. Between 113 and 127 the 13 consecutive integers are divisible by 2,5,2,3,2,7,2,11,2,3,2,5,2, each from {2,3,5,7,11}. Thus a(5)=114, the smallest with this property.


CROSSREFS

Cf. A002110, A048670.
Sequence in context: A084954 A226281 A217993 * A339017 A084957 A239944
Adjacent sequences: A049297 A049298 A049299 * A049301 A049302 A049303


KEYWORD

hard,nonn


AUTHOR

Labos Elemer


EXTENSIONS

More terms from T. D. Noe, Mar 29 2007
a(11)a(12) from Donovan Johnson, Oct 13 2009
a(13) from Donovan Johnson, Oct 20 2009
Terms a(14)a(24) from Max Alekseyev, Nov 14 2009


STATUS

approved



