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A188640
Decimal expansion of e + sqrt(1+e^2).
62
5, 6, 1, 4, 6, 6, 8, 5, 6, 0, 0, 4, 9, 0, 5, 3, 4, 3, 9, 2, 5, 4, 7, 8, 2, 8, 3, 3, 1, 8, 6, 3, 3, 7, 3, 6, 0, 2, 3, 9, 8, 2, 0, 5, 6, 4, 1, 7, 1, 1, 3, 3, 9, 9, 6, 3, 2, 0, 4, 7, 8, 1, 4, 6, 4, 7, 2, 9, 3, 9, 2, 5, 6, 4, 2, 3, 9, 0, 0, 2, 6, 5, 0, 9, 8, 0, 4, 8, 4, 2, 8, 5, 5, 3, 4, 1, 5, 3, 5, 1, 3, 3, 7, 3, 7, 6, 0, 7, 6, 8, 8, 0, 8, 7, 8, 3, 3, 6, 0, 7, 7, 0, 0, 4, 0, 1, 8, 2, 9, 9
OFFSET
1,1
COMMENTS
The shape of a rectangle WXYZ, denoted by [WXYZ], is defined by length/width: [WXYZ]=max{|WX|/|YZ|, |YZ|/|WX|}. Consider the following configuration of rectangles AEFD, EBCF, ABCD, where AEFD is not a square:
D................F....C
.......................
.......................
.......................
A................E....B
Let r=[AEFD]. The r-extension rectangle of AEFD is here introduced as the rectangle ABCD for which [AEFD]=[EBCF] and |AE|<>|EB|. That is, AEFD has the prescribed shape r, and AEFD and EBCF are similar without being congruent.
We extend the definition of r-extension rectangle to the case that 0<r<1; in this case, [AEFD]=1/r and ABCD is defined as above.
Then for all r>0, it is easy to prove that [ABCD] = (r+sqrt(4+r^2))/2.
This here is the length/width ratio for the (2e)-extension rectangle.
A (2e)-extension rectangle matches the continued fraction A188796 for the shape L/W=(e+sqrt(1+e^2). This is analogous to the matching of a golden rectangle to the continued fraction [1,1,1,1,1,1,1,...]. Specifically, for the (2e)-extension rectangle, 5 squares are removed first, then 1 square, then 1 square, then 1 square, then 1 square, then 2 squares..., so that the original rectangle is partitioned into an infinite collection of squares.
Shapes of other r-extension rectangles, partitionable into a collection of squares in accord with the continued fraction of the shape [ABCD], are approximated at A188635-A188639, A188655-A188659, and A188720-A188737.
For (related) r-contraction rectangles, see A188738 and A188739.
LINKS
Clark Kimberling, A Visual Euclidean Algorithm, The Mathematics Teacher 76 (1983) 108-109.
Clark Kimberling, Two kinds of golden triangles, generalized to match continued fractions, Journal for Geometry and Graphics, 11 (2007) 165-171.
FORMULA
Equals exp(A366599). - Amiram Eldar, Oct 18 2023
EXAMPLE
Length/width = 5.61466856004905343925478283318633736023982...
MAPLE
evalf(exp(1)+sqrt(1+exp(2)), 140); # Muniru A Asiru, Nov 01 2018
MATHEMATICA
r=2E; t=(r+(4+r^2)^(1/2))/2; FullSimplify[t]
N[t, 130]
RealDigits[N[t, 130]][[1]]
PROG
(PARI) exp(1)+sqrt(1+exp(2)) \\ Charles R Greathouse IV, Jun 16 2011
(Magma) SetDefaultRealField(RealField(100)); Exp(1) + Sqrt(1 + Exp(2)); // G. C. Greubel, Oct 31 2018
CROSSREFS
KEYWORD
nonn,cons,easy
AUTHOR
Clark Kimberling, Apr 10 2011
STATUS
approved