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Search: id:A141467
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| A141467 |
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a(1)=1, else transformed product of prime factors of the n-th composite, the largest prime incremented by 3, the smallest decremented by 1. |
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+0
1
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| 1, 6, 10, 12, 8, 12, 10, 16, 20, 18, 16, 20, 14, 24, 32, 16, 36, 20, 24, 40, 28, 20, 40, 36, 22, 32, 32, 30, 28, 48, 26, 48, 60, 40, 40, 32, 54, 56, 40, 44, 32, 48, 34, 60, 80, 64, 42, 40, 52, 50, 72, 40, 80, 44, 84, 48, 64, 108, 44, 60, 80, 46, 64, 56, 72, 96, 52, 68, 50, 88, 96, 70, 84
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OFFSET
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1,2
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COMMENT
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In the prime number decomposition of k=A002808(n), one instance of the largest prime, pmax=A052369(n), is replaced by pmax+3, and one instance of the smallest prime, pmin=A056608(n), is replaced by pmin-1. a(n) is the product of this modified set of factors if nonprime. The case of n=1, k=4, is the only case where this modified product (2+3)*(2-1)=5 is prime, and listed as a(1)=1.
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FORMULA
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a(n)=k(pmax+3)(pmin-1)/(pmin*pmax), n>1, where k=A002808(n), pmin=A056608(n), pmax=A052369(n).
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EXAMPLE
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If k(1)=4=(p(max)=2)*(p(min)=2), then (2+3)*(2-1)=5*1=5 (prime).
If k(2)=6=(p(max)=3)*(p(min)=2), then (3+3)*(2-1)=6*1=6=a(2).
If k(3)=8=(p(max)=2)*(p=2)*(p(min)=2), then (2+3)*2*(2-1)=5*2*1=10=a(3).
If k(4)=9=(p(max)=3)*(p(min)=3), then (3+3)*(3-1)=6*2=12=a(4).
If k(5)=10=(p(max)=5)*(p(min)=2), then (5+3)*(2-1)=8*1=8=a(5).
If k(6)=12=(p(max)=3)*(p=2)*(p(min)=2), then (3+3)*2*(2-1)=6*2*1=12=a(6).
If k(7)=14=(p(max)=7)*(p(min)=2), then (7+3)*(2-1)=10*1=10=a(7).
If k(8)=15=(p(max)=5)*(p(min)=3), then (5+3)*(3-1)=8*2=16=a(8),
If k(9)=16=(p(max)=2)*2*2*(p(min)=2), then (2+3)*2*2*(2-1)=5*2*2*1=20=a(9).
If k(10)=18=(p(max)=3)*(p=3)*(p(min)=2), then (3+3)*3*(2-1)=6*3*1=18=a(10), etc.
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CROSSREFS
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Sequence in context: A136812 A109397 A133210 this_sequence A105642 A064040 A024619
Adjacent sequences: A141464 A141465 A141466 this_sequence A141468 A141469 A141470
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KEYWORD
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nonn
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AUTHOR
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Juri-Stepan Gerasimov (2stepan(AT)rambler.ru) Aug 08 2008
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EXTENSIONS
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Edited by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Aug 14 2008
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