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A197594
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Sum of the cubes of the first odd numbers up to a(n) equals the n-th perfect number.
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0
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3, 7, 15, 127, 511, 1023, 65535, 2147483647, 35184372088831, 18014398509481983, 18446744073709551615, 3705346855594118253554271520278013051304639509300498049262642688253220148477951
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OFFSET
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2,1
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COMMENTS
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Except for the first perfect number 6, every even perfect number 2^(p-1)*(2^p - 1) is the sum of the cubes of the first 2^((p-1)/2) odd numbers.
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REFERENCES
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Albert H. Beiler: Recreations in the theory of numbers, New York, Dover, Second Edition, 1966, p. 22.
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LINKS
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FORMULA
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1/8*(a(n) + 1)^2*(a(n)^2 + 2*a(n) - 1) = 2^(p-1)*(2^p - 1) with p = 2*log(a(n) + 1)/log(2) - 1 a Mersenne prime.
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EXAMPLE
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a(2)=3, since 1^3 + 3^3 = 28, which is the second perfect number.
a(3)=7, since 1^3 + 3^3 + 5^3 + 7^3 = 496, which is the third perfect number.
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CROSSREFS
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KEYWORD
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nonn,changed
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AUTHOR
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STATUS
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approved
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