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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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0
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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, 31, 32, 3, 8, 2, 15, 5, 2, 4, 5, 37, 2, 17, 3, 10, 5, 3, 41, 2, 1, 4, 43, 4, 7, 5, 4, 2, 21, 47, 3, 13, 49, 2, 23, 3, 14, 4, 9, 53, 2, 25, 5, 6, 7, 1, 3, 16, 2
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Row n contains A001221(n) terms.
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EXAMPLE
| 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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PROG
| (PARI) { a(n) = local(f); if(n==1, return([1])); f=factorint(n); f=vecsort(vector(matsize(f)[1], i, f[i, 1]^f[i, 2])); vector(#f, i, f[i]-if(i>1, f[i-1])) } concat(vector(100, n, a(n))) [From Max Alekseyev (maxale(AT)gmail.com), May 07 2009]
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CROSSREFS
| Cf. A141810, A001221.
Sequence in context: A037901 A037839 A165072 * A070671 A119281 A173525
Adjacent sequences: A141401 A141402 A141403 * A141405 A141406 A141407
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KEYWORD
| nonn,tabf
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AUTHOR
| Leroy Quet Aug 03 2008
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EXTENSIONS
| Extended by Max Alekseyev (maxale(AT)gmail.com), May 07 2009
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