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 A069929 Number of k, 1<=k<=n, such that k^3+1 divides n^3+1. 0
 1, 1, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 4, 1, 3, 2, 2, 1, 3, 1, 2, 3, 3, 1, 3, 1, 3, 2, 2, 1, 3, 1, 2, 2, 2, 1, 4, 1, 2, 2, 2, 1, 5, 1, 3, 2, 2, 1, 3, 1, 3, 2, 2, 1, 5, 1, 3, 2, 2, 2, 3, 1, 2, 3, 4, 1, 3, 1, 2, 2, 4, 1, 3, 1, 2, 2, 2, 1, 5, 1, 2, 2, 3, 1, 4, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 4, 1, 4, 2, 2 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 LINKS FORMULA Conjecture: (1/n)*Sum_{k=1..n} a(k) = C*log(log(n)) + o(log(log(n)) with 1 < C < 3/2. PROG (PARI) for(n=1, 150, print1(sum(i=1, n, if((n^3+1)%(i^3+1), 0, 1)), ", ")) CROSSREFS Cf. A066743. Sequence in context: A101872 A251659 A174892 * A101312 A241273 A154263 Adjacent sequences:  A069926 A069927 A069928 * A069930 A069931 A069932 KEYWORD easy,nonn AUTHOR Benoit Cloitre, May 05 2002 STATUS approved

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