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Search: id:A164799
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| A164799 |
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a(n) = the smallest positive integer such that the product of a(n) consecutive integers, where n is the smallest, is divisible by every prime from 2 to the largest prime divisor of the product. (a(1)=1.) |
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+0
4
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| 1, 1, 2, 1, 2, 1, 4, 1, 2, 5, 5, 1, 10
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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a(n) = A164798(n) - n +1.
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EXAMPLE
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Consider the products of consecutive integers, (m+9)!/9!, m >= 1. First, 10 is divisible by 2 and 5, but there is a prime gap since 3 is missing from the factorization. 10*11 is divisible by 2, 5, and 11, but 3 and 7 are missing. 10*11*12 is divisible by 2, 3, 5, and 11, but 7 is missing. 10*11*12*13 is divisible by all primes up to 13, except 7. But 10*11*12*13*14 is indeed divisible by every prime from 2 to 13. So, a(10) = 5 because 5 consecutive numbers are multiplied together.
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CROSSREFS
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A164798
Sequence in context: A071222 A067005 A135517 this_sequence A072614 A067044 A055684
Adjacent sequences: A164796 A164797 A164798 this_sequence A164800 A164801 A164802
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KEYWORD
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more,nonn
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AUTHOR
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Leroy Quet (q1qq2qqq3qqqq(AT)yahoo.com), Aug 26 2009
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