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Search: id:A140449
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| A140449 |
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a(n) = the multiple of A142972(n) such that the nth prime <= a(n) <= the (n+1)th prime. |
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1
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| 3, 5, 6, 10, 12, 15, 18, 20, 27, 30, 36, 40, 42, 45, 48, 54, 60, 63, 70, 72, 77, 80, 88, 90, 100, 102, 105, 108, 110, 120, 130, 135, 138, 140, 150, 153, 160, 165, 170, 176, 180, 187, 192, 195, 198, 204, 221, 225, 228, 230, 238, 240, 242, 252, 259, 264, 270, 272
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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There is always only one multiple of A142972(n) that is between the nth prime and the (n+1)th prime.
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LINKS
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Leroy Quet, Home Page (listed in lieu of email address)
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EXAMPLE
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The 15th prime is 47 and the 16th prime is 53. So we will examine the integers 47,48,49,50,51,52,53. Now, 1 divides each of these integers. 2 divides 48,50,52. 3 divides 48 and 51. 4 divides 48 and 52. 5 divides 50. 6 divides 48. 7 divides 49. 8 divides 48. But 9 doesn't divide any integer in the span of consecutive integers 47 to 53. So 8 is the largest integer m such that 1,2,3,4,...m each divide at least one integer in the span 47 to 53. 48 is the multiple of 8 among the integers in the span. So a(15) = 48.
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CROSSREFS
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Cf. A142972.
Sequence in context: A028927 A099190 A122772 this_sequence A115823 A112926 A027627
Adjacent sequences: A140446 A140447 A140448 this_sequence A140450 A140451 A140452
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KEYWORD
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nonn
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AUTHOR
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Leroy Quet Jul 21 2008
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EXTENSIONS
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Extended by Ray Chandler (rayjchandler(AT)sbcglobal.net), Jun 21 2009
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