OFFSET
1,1
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.
REFERENCES
J.-M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique Des Nombres, Problem 261 pp. 186, Ellipses, Paris 2004.
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.
EXAMPLE
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.
PROG
(PARI) isok(n) = if (!ispower(n, 3), my(f=factor(n)); sum(k=1, #f~, f[k, 1]^3) == n); \\ Michel Marcus, Dec 25 2018
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Naohiro Nomoto, Nov 21 2000
EXTENSIONS
Corrected definition and five more terms from Koffie Duah (admc1961(AT)live.com), Feb 16 2008
STATUS
approved