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Search: id:A049300
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| A049300 |
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Smallest number which begins the maximal number of consecutive integers divisible by one of the first n prime numbers. |
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+0 2
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| 2, 2, 2, 2, 114, 9440, 217128, 60044, 20332472, 417086648
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OFFSET
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1,1
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COMMENT
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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
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FORMULA
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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
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EXAMPLE
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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.
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CROSSREFS
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Cf. A002110, A048670.
Sequence in context: A029627 A075182 A084954 this_sequence A084957 A035307 A004481
Adjacent sequences: A049297 A049298 A049299 this_sequence A049301 A049302 A049303
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KEYWORD
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hard,nonn
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu)
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EXTENSIONS
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More terms from T. D. Noe (noe(AT)sspectra.com), Mar 29 2007
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