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A317969
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Decimal expansion of (2^(1/3)-1)^(1/3).
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1
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6, 3, 8, 1, 8, 5, 8, 2, 0, 8, 6, 0, 6, 4, 4, 1, 5, 3, 0, 1, 5, 5, 0, 3, 6, 5, 9, 4, 4, 4, 0, 6, 7, 7, 0, 1, 2, 6, 5, 1, 5, 7, 5, 4, 3, 9, 7, 7, 9, 9, 7, 6, 8, 3, 4, 2, 1, 0, 6, 2, 0, 8, 1, 5, 8, 0, 5, 7, 5, 4, 8, 5, 1, 3, 9, 7, 0, 7, 9, 2, 5, 0, 2, 7, 6
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OFFSET
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0,1
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COMMENTS
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(2^(1/3)-1)^(1/3) = (1/9)^(1/3) - (2/9)^(1/3) + (4/9)^(1/3) is a famous and remarkable identity of Ramanujan's.
Ramanujan's question 1076 (ii), see Berndt and Rankin in References: Show that (4*(2/3)^(1/3)-5*(1/3)^(1/3))^(1/8) = (4/9)^(1/3)-(2/9)^(1/3)+(1/9)^(1/3). - Hugo Pfoertner, Aug 28 2018
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REFERENCES
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B. C. Berndt and R. A. Rankin, Ramanujan: Essays and Surveys, American Mathematical Society, 2001, ISBN 0-8218-2624-7, page 222 (JIMS 11, page 199).
S. Ramanujan, Coll. Papers, Chelsea, 1962, page 331, Question 682; page 334 Question 1076.
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LINKS
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Susan Landau, Simplification of nested radicals, SIAM Journal on Computing 21.1 (1992): 85-110. See page 85. [Do not confuse this paper with the short FOCS conference paper with the same title, which is only a few pages long.]
S. Ramanujan, Question 682, Journal of the Indian Mathematical Society, VII, p. 160.
S. Ramanujan, Question 1076, Journal of the Indian Mathematical Society, XI, p. 199.
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FORMULA
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k^(3*n) = x(n) + A002580*y(n) + A005480*z(n) where k is this constant z(n) = A108369(n-1), y(n) = z(n)+z(n+1), x(n) = y(n)+y(n+1); A002580 and A005480 are the cube root of 2 and 4. (End)
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EXAMPLE
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0.638185820860644153015503659444067701265157543977997683421...
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MAPLE
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evalf((4*(2/3)^(1/3)-5*(1/3)^(1/3))^(1/8)); # Muniru A Asiru, Aug 28 2018
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MATHEMATICA
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PROG
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(PARI) (4*(2/3)^(1/3)-5*(1/3)^(1/3))^(1/8) /* Hugo Pfoertner Aug 28 2018 */
(PARI) sqrtn(1/9, 3) - sqrtn(2/9, 3) + sqrtn(4/9, 3) \\ Michel Marcus, Jan 07 2022
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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