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Start with x=4/3; repeatedly apply the map x -> (x^2) ceiling(x); sequence gives numerators of the resulting sequence of fractions.
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%I #11 Jun 16 2016 03:12:23

%S 4,32,4096,285212672,3536203627938199896064,

%T 27735467127437590594631628902073909856749798039036448735232

%N Start with x=4/3; repeatedly apply the map x -> (x^2) ceiling(x); sequence gives numerators of the resulting sequence of fractions.

%C In this approximate cubing, does an iteration eventually yield an integer, after which denominators are 1? Fractions are 4/3, 32/9, 4096/81, 285212672/2187, 3536203627938199896064/1594323, 27735467127437590594631628902073909856749798039036448735232/2541865828329, 8393707510592229745861012598171776416393703955772365464679357805492895042198412632866136478758067686243059846017657263750451410617880163800261945260539460460740608/6461081889226673298932241.

%C a(9) has 1343 digits, and is too large for a b-file. - _Robert Israel_, Jun 15 2016

%H Robert Israel, <a href="/A117620/b117620.txt">Table of n, a(n) for n = 1..8</a>

%H J. C. Lagarias and N. J. A. Sloane, Approximate squaring (<a href="http://neilsloane.com/doc/apsq.pdf">pdf</a>, <a href="http://neilsloane.com/doc/apsq.ps">ps</a>), Experimental Math., 13 (2004), 113-128.

%e a(4) = 285212672 because (4096/81)^2 * ceiling(4096/81) = (4096/81)^2 * ceiling(4096/81) = * ceiling(50.5679012) = (16777216/6561) * 51 = 285212672/2187.

%p x[1]:= 4/3:

%p for n from 1 to 9 do x[n+1]:= x[n]^2*ceil(x[n]) od:

%p seq(numer(x[i]),i=1..10); # _Robert Israel_, Jun 15 2016

%Y Cf. A072340, A085276, A117596.

%K easy,frac,nonn

%O 1,1

%A _Jonathan Vos Post_, Apr 07 2006

%E Erroneous term removed by _Giovanni Resta_, Jun 15 2016