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A329732
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a(n) is the smallest m for which there is a sequence n = b_1 < b_2 < ... < b_t = m such that b_1*b_2*...*b_t is a perfect cube.
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3
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0, 1, 4, 9, 9, 18, 18, 21, 8, 16, 24, 33, 18, 39, 28, 30, 25, 51, 25, 57, 36, 36, 44, 69, 42, 36, 52, 27, 45, 87, 45, 93, 49, 55, 68, 60, 48, 111, 76, 65, 60, 123, 54, 129, 66, 54, 92, 141, 70, 56, 72, 85, 78, 159, 80, 80, 84, 95, 116, 177, 84, 183, 124, 84, 64
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OFFSET
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0,3
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COMMENTS
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For each k there exists a sufficiently large N such that for all primes p > N, a(k*p) = (k+2)*p. [We can prove the proposition is true for N = 64*(t*k)^2, where t = k*(k+1)*(k+2): there is a positive integer x such that t^2*x^3 < k*p < t^2*(x+1)^3 < t^2*x^3*(k+1)/k < (k+1)*p for p > N. So one increasing sequence starting with k*p, ending with (k+2)*p, and having a product which is a perfect cube is (k*p) * (t^2*(x+1)^3) * ((k+1)*p) * ((k+2)*p) = (t*p*(x+1))^3. Noticed that a(k*p) >= (k+2)*p (because b_1*b_2*...*b_t is divisible by p^3) for p > N, so a(k*p) = (k+2)*p. - Jinyuan Wang, Dec 22 2021]
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LINKS
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FORMULA
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a(p) = 3*p for all primes p >= 7.
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EXAMPLE
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For n = 22, one increasing sequence starting with 22, ending with a(22) = 44, and having a product which is a perfect cube is 22 * 24 * 25 * 30 * 32 * 33 * 44 = 2640^3.
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CROSSREFS
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A cube analog of R. L. Graham's sequence (A006255).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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