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%I #19 Sep 20 2023 05:50:21
%S 13,203,1615,51595,412529,6599099,52788535,3378355987,27026481101,
%T 432421205841,3459361042977,110699432952143,885595037556565,
%U 14169517557800915,113356129507566775,14509583941597490435
%N Numerators of partial sums of a series for sqrt(2) + sqrt(3) involving Catalan numbers.
%C The corresponding denominators are 4*A120785(n).
%C Sqrt(2)+sqrt(3) = (4*sin(Pi/4) + 6*tan(Pi/6))/2 = 3.146264370 (maple10, 10 digits). This is the arithmetic mean of the areas of an 8-gon (octagon), resp. 6-gon (hexagon) inscribed, resp. circumscribed in a unit circle.
%C Popper (see the reference) argues that Plato knew about the sum of sqrt(2)+sqrt(3). This sum approximates Pi with a relative error of 0.15%. The two right triangles, one with side lengths (1,1/2,sqrt(3)/2) and the other with side lengths (sqrt(2),1,1) are used in Plato's Timaios [53d] to build four of the five regular polyhedra (Platonic solids).
%C The Taylor series for sqrt(2) = sqrt(1+1) and sqrt(3) = 3*sqrt(1-2/3) are used here. Therefore lim_{n->oo} r(n) = sqrt(2)+sqrt(3), with rationals r(n) defined below.
%D K. R. Popper, Die Welt des Parmenides, Piper, 2001, 2005. Ch. 8: Platon und die Geometrie (1950), pp. 326-337. English: The World of Parmenides, Routledge, London, New York, 1998.
%H Wolfdieter Lang, <a href="/A121503/a121503.txt">Rationals r(n), limit</a>.
%F a(n)= numerator(r(n)) with r(n):= 4-(sum(C(k)*(1+2^(k+1))/16^k,k=0..n)/4, n>=0, with C(k)=A000108(k) (Catalan numbers).
%e Rationals r(n): [13/4, 203/64, 1615/512, 51595/16384, 412529/131072, 6599099/2097152, 52788535/16777216,...].
%o (PARI) a(n) = numerator(4 - sum(k=0, n, binomial(2*k,k)/(k+1)*(1+2^(k+1))/16^k)/4); \\ _Michel Marcus_, Sep 20 2023
%Y Cf. A000108, A120785.
%K nonn,easy,frac
%O 0,1
%A _Wolfdieter Lang_, Aug 16 2006
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