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A122140
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Numbers m that divide the sum of cubes of the first m primes A098999(m).
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5
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1, 25, 537, 661, 5199, 113253, 240621, 5337048977, 17434578479, 34216676921, 1991831965911, 4495321247369, 22567781434431, 37328883555791, 110447613624133, 188368390470877, 324587968952249, 1983705516917661
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OFFSET
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1,2
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COMMENTS
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LINKS
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EXAMPLE
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a(2) = 25 because 25 is the first number n>1 that divides sum of cubes of the first n primes. A098999(25) mod 25 = (2^3 + 3^3 + 5^3 + ... + 89^3 + 97^3) mod 25 = 0.
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MATHEMATICA
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s = 0; t = {}; Do[s = s + Prime[n]^3; If[ Mod[s, n] == 0, AppendTo[t, n]], {n, 1000000}]; t
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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