

A317969


Decimal expansion of (2^(1/3)1)^(1/3).


1



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

0,1


COMMENTS

(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


REFERENCES

B. C. Berndt and R. A. Rankin, Ramanujan: Essays and Surveys, American Mathematical Society, 2001, ISBN 0821826247, page 222 (JIMS 11, page 199).
S. Ramanujan, Coll. Papers, Chelsea, 1962, page 331, Question 682; page 334 Question 1076.


LINKS

Table of n, a(n) for n=0..84.
Susan Landau, Simplification of nested radicals, SIAM Journal on Computing 21.1 (1992): 85110. 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.]


EXAMPLE

0.638185820860644153015503659444067701265157543977997683421...


MAPLE

evalf((4*(2/3)^(1/3)5*(1/3)^(1/3))^(1/8)); # Muniru A Asiru, Aug 28 2018


PROG

(PARI) (4*(2/3)^(1/3)5*(1/3)^(1/3))^(1/8) /* Hugo Pfoertner Aug 28 2018 */


CROSSREFS

Cf. A318526.
Sequence in context: A182566 A115287 A061533 * A281682 A157294 A021098
Adjacent sequences: A317966 A317967 A317968 * A317970 A317971 A317972


KEYWORD

nonn,cons


AUTHOR

N. J. A. Sloane, Aug 27 2018


STATUS

approved



