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A105548
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Continued fraction expansion of prime nested radical A105546.
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5
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2, 9, 1, 1, 1, 7, 3, 5, 4, 1, 1, 1, 3, 2, 2, 1, 1, 1, 1, 15, 1, 3, 1, 41, 6, 1, 3, 1, 3, 10, 1, 1, 1, 9, 9, 1, 25, 1, 3, 1, 1, 2, 2, 2, 1, 34, 59, 2, 2, 2, 1, 2, 2, 3, 3, 1, 5, 2, 21, 3, 4, 10, 1, 3, 20, 2, 3, 2, 1, 4, 7, 1, 6, 1, 6, 3, 4, 1, 3, 5, 6, 1, 1, 4, 1, 3, 6, 25, 7, 2, 1, 1, 2, 1, 6, 1, 1, 7, 1, 3, 2
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OFFSET
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1,1
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COMMENTS
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Records are: 9,15,41,59,117,153,599,1663,8212,..., . Robert G. Wilson v: "It would appear superficially that this constant is normal." Sqrt(1 + sqrt(2 + sqrt(3 + sqrt(4 + ... = ~ 1.75793275661800... "It was discovered by T. Vijayaraghavan that the infinite radical, sqrt( a_1 + sqrt( a_2 + sqrt ( a_3 + sqrt( a_4 + ... where a_n => 0, will converge to a limit if and only if the limit of (ln a_n)/2^n exists." [Clawson, 229; A072449] We know the asymptotic limit of primes and hence that the Prime Nested Radical converges.
Clawson misstates Vijayaraghavan's theorem. Vijayaraghavan proved that for a_n > 0, the infinite radical sqrt(a_1 + sqrt(a_2 + sqrt(a_3 + ...))) converges if and only if limsup (log a_n)/2^n < infinity. (For example, suppose a_n = 1 if n is odd, and a_n = e^2^n if n is even. Then (log a_n)/2^n = 0, 1, 0, 1, 0, 1, ... for n >= 1, so the limit does not exist. However, limsup (log a_n)/2^n = 1 and the infinite radical converges.) - Jonathan Sondow, Mar 25 2014
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REFERENCES
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Calvin C. Clawson, "Mathematical Mysteries, the beauty and magic of numbers," Perseus Books, Cambridge, Mass., 1996, pages 142 & 229.
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LINKS
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Jonathan M. Borwein and G. de Barra, Nested Radicals, Amer. Math. Monthly 98, 735-739, 1991.
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FORMULA
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Continued fraction expansion of sqrt(2 + sqrt(3 + sqrt(5 + sqrt(7 + sqrt(11 + ... + sqrt(prime(n))))).
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MATHEMATICA
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f[n_] := Block[{k = n, s = 0}, While[k > 0, s = Sqrt[s + k]; k-- ]; s]; ContinuedFraction[ f[100], 101]; (* Robert G. Wilson v *)
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CROSSREFS
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KEYWORD
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cofr,easy,nonn
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AUTHOR
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STATUS
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approved
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