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A058041
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Noncubes equal to the sum of cubes of their prime factors.
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0
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OFFSET
| 1,1
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COMMENTS
| Let (p_1,p_2,...,p_m) = prime factors of n (m>1, k>0); sequence gives n such that n = Sum_{i=1,2,..,m} (p_i)^3.
Repeated prime factors are used only once.
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REFERENCES
| J.-M. De Koninck and A. Mercier, 1001 Problemes en Theorie Classique Des Nombres, Problem 261 pp. 186, Ellipses, Paris 2004.
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LINKS
| Jean-Marie De Koninck, Partial Sums of Powers of Prime Factors, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.6
J. M. de Koninck and Armel Mercier, 1001 Problems in Classical Number Theory, American Mathematical Society 2007. Page 160.
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EXAMPLE
| 378=2*3^3*7=2^3+3^3+7^3 (k=3); 2548=2^2*7^2*13=2^3+7^3+13^3 (k=3); 2836295=5*7*11*53*139=5^3+7^3+11^3+53^3+139^3 (k=3);
378 = 2 * 3^3 * 7 = 2^3 + 3^3 + 7^3
2548 = 2^2 * 7^2 * 13 = 2^3 + 7^3 + 13^3
2836295 = 5 * 7 * 11 * 53 * 139 = 5^3 + 7^3 + 11^3 + 53^3 + 139^3
4473671462 = 2 * 13 * 179 * 593 * 1621 = 2^3 + 13^3 + 179^3 + 593^3 + 1621^3
23040925705 = 5 * 7 * 167 * 1453 * 2713 = 5^3 + 7^3 + 167^3 + 1453^3 + 2713^3
13579716377989 = 19 * 157 * 173 * 1103 * 23857 = 19^3 + 157^3 + 173^3 + 1103^3 + 23857^3
21467102506955 = 5 * 73 * 313 * 1439 * 27791 = 5^3 + 7^3 + 313^3 + 1439^3 + 27791^3
119429556097859 = 7 * 53 * 937 * 6983 * 49199 = 7^3 + 53^3 + 937^3 + 6983^3 + 49199^3
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CROSSREFS
| Sequence in context: A132648 A157116 A071624 * A154078 A047632 A033699
Adjacent sequences: A058038 A058039 A058040 * A058042 A058043 A058044
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KEYWORD
| nonn,more
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AUTHOR
| Naohiro Nomoto (6284968128(AT)geocities.co.jp), Nov 21 2000
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EXTENSIONS
| Corrected definition and five more terms from Koffie Duah (admc1961(AT)live.com), Feb 16 2008
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