OFFSET
0,1
COMMENTS
The corresponding denominators are 4*A120785(n).
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.
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).
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.
REFERENCES
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.
LINKS
Wolfdieter Lang, Rationals r(n), limit.
FORMULA
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).
EXAMPLE
Rationals r(n): [13/4, 203/64, 1615/512, 51595/16384, 412529/131072, 6599099/2097152, 52788535/16777216,...].
PROG
(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
CROSSREFS
KEYWORD
nonn,easy,frac
AUTHOR
Wolfdieter Lang, Aug 16 2006
STATUS
approved