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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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2
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2, 2, 2, 2, 114, 9440, 217128, 60044, 20332472, 417086648, 74959204292, 187219155594, 79622514581574, 14478292443584, 6002108856728918, 12288083384384462, 5814429911995661690, 14719192159220252523420
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| 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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LINKS
| Max Alekseyev, Table of n, a(n) for n=1..24 [From Max Alekseyev (maxale(AT)gmail.com), Nov 15 2009]
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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
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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.
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CROSSREFS
| Cf. A002110, A048670.
Sequence in context: A075182 A154288 A084954 * A084957 A035307 A004481
Adjacent sequences: A049297 A049298 A049299 * A049301 A049302 A049303
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KEYWORD
| hard,nonn
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AUTHOR
| Labos E. (labos(AT)ana.sote.hu)
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EXTENSIONS
| More terms from T. D. Noe (noe(AT)sspectra.com), Mar 29 2007
a(11)-a(12) from Donovan Johnson (donovan.johnson(AT)yahoo.com), Oct 13 2009
a(13) from Donovan Johnson (donovan.johnson(AT)yahoo.com), Oct 20 2009
Terms a(14) onwards from Max Alekseyev (maxale(AT)gmail.com), Nov 14 2009
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