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A049300 Smallest number which begins the maximal number of consecutive integers divisible by one of the first n prime numbers. +0
2
2, 2, 2, 2, 114, 9440, 217128, 60044, 20332472, 417086648 (list; graph; listen)
OFFSET

1,1

COMMENT

The length of these chains is given by the first maximal gaps minus 1 in reduced residue systems of consecutive primorial numbers: 1,1,3,5,9,13,21,25, etc. (A048670 - 1).

Let j(m) be the Jacobsthal function (A048669): 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 (noe(AT)sspectra.com), Mar 29 2007

FORMULA

One of prime(1), ..., prime(n) divides a maximal number of consecutive integers starting with a(n), which is minimal of this property.

a(n)=1+A128707(A002110(n)) - T. D. Noe (noe(AT)sspectra.com), 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: A029627 A075182 A084954 this_sequence A084957 A035307 A004481

Adjacent sequences: A049297 A049298 A049299 this_sequence A049301 A049302 A049303

KEYWORD

hard,nonn

AUTHOR

Labos E. (labos(AT)ana.sote.hu)

EXTENSIONS

More terms from T. D. Noe (noe(AT)sspectra.com), Mar 29 2007

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Last modified August 19 23:53 EDT 2008. Contains 142930 sequences.


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