



2, 5, 8, 4, 4, 0, 1, 7, 2, 4, 0, 1, 9, 7, 7, 6, 7, 2, 4, 8, 1, 2, 0, 7, 6, 1, 4, 7, 1, 5, 3, 3, 3, 1, 3, 4, 2, 1, 1, 2, 3, 8, 2, 0, 9, 0, 4, 6, 7, 9, 6, 9, 0, 0, 0, 3, 1, 3, 4, 3, 8, 5, 8, 3, 9, 6, 7, 5, 4, 4, 8, 2, 9, 8, 9, 1, 8, 6, 7, 9, 6, 3, 6, 1, 4, 0, 8, 8, 7, 4, 6, 9, 7, 7, 8, 0, 1, 8, 6, 9, 6, 4, 2, 7, 2
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OFFSET

1,1


COMMENTS

A112373 is defined by the recurrence: let b(n) = A112373(n), then
b(n) =(b(n1)^3 + b(n1)^2)/b(n2) for n>=2 with b(0)=b(1)=1.
Thus the sum of unit fractions 1/A112373(n) converges rapidly.


LINKS

Table of n, a(n) for n=1..105.
Andrew N. W. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbers, arXiv:1507.00063 [math.NT], 2015.


EXAMPLE

2.584401724019776724812076147153331342112382090467969...
= Sum_{n>=0} 1/A112373(n) = 1/1 +1/1 +1/2 +1/12 +1/936 +1/68408496 +...
= [2;1,1,2,2,6,12,78,936,73086,68408496,...] (continued fraction).


CROSSREFS

Cf. A112373, A114551 (continued fraction), A114552.
Sequence in context: A075175 A075173 A163337 * A094001 A020859 A062089
Adjacent sequences: A114547 A114548 A114549 * A114551 A114552 A114553


KEYWORD

cons,nonn


AUTHOR

Paul D. Hanna, Dec 08 2005


STATUS

approved



