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Search: id:A141404
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| A141404 |
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Irregular array: For any prime p that divides n, if the highest power of the prime p that divides n is p^b(n,p), then p^b(n,p) = sum{k=1 to m} a(n,k), where m is the order of the prime-power p^b(n,p) among the prime-powers (each being the highest power of each prime q that divides n, where q divides n) when they are ordered by size. Row 1 = (1). |
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1
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| 1, 2, 3, 4, 5, 2, 1, 7, 8, 9, 2, 3, 11, 3, 1, 13, 2, 5, 3, 2, 16, 17, 2, 7, 19, 4, 1, 3, 4, 2, 9, 23, 3, 5, 25, 2, 11, 27, 4, 3, 29, 2, 1, 2
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Row n contains A001221(n) terms.
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EXAMPLE
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The prime-factorization of 300 is 2^2 *3^1 *5^2. So the prime-powers ordered by size are: 3,4,25. Therefore row 300 is (3,1,21), because 3=3, 3+1 = 4, 3+1+21 = 25.
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CROSSREFS
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Cf. A141810, A001221.
Sequence in context: A037893 A037901 A037839 this_sequence A070671 A119281 A070772
Adjacent sequences: A141401 A141402 A141403 this_sequence A141405 A141406 A141407
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KEYWORD
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more,nonn,tabf
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AUTHOR
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Leroy Quet (q1qq2qqq3qqqq(AT)yahoo.com), Aug 03 2008
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