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A069929
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Number of k, 1 <= k <= n, such that k^3+1 divides n^3+1.
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2
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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
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OFFSET
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1,3
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LINKS
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FORMULA
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Conjecture: (1/n)*Sum_{k=1..n} a(k) = C*log(log(n)) + o(log(log(n)) with 1 < C < 3/2.
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EXAMPLE
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a(5) = 3 because among the numbers 1^3+1 = 2, 2^3+1 = 9, 3^3+1 = 28, 4^3+1 = 65, and 5^3 + 1 = 126, only 3 of them (2, 9, 126) divide 5^3+1 = 126. - Petros Hadjicostas, Sep 18 2019
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MAPLE
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a:= n-> add(`if`(irem(n^3+1, k^3+1)=0, 1, 0), k=1..n):
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PROG
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(PARI) for(n=1, 150, print1(sum(i=1, n, if((n^3+1)%(i^3+1), 0, 1)), ", "))
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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