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A188738
Decimal expansion of e-sqrt(e^2-1).
20
1, 9, 0, 6, 2, 3, 6, 0, 4, 1, 4, 7, 3, 3, 0, 6, 1, 4, 2, 5, 9, 4, 2, 8, 2, 5, 6, 5, 4, 1, 5, 5, 5, 2, 6, 8, 6, 6, 3, 0, 2, 2, 2, 0, 2, 0, 9, 8, 3, 5, 6, 4, 6, 1, 7, 3, 5, 2, 7, 3, 3, 7, 6, 8, 0, 9, 7, 0, 9, 0, 8, 8, 4, 4, 9, 2, 2, 1, 1, 4, 1, 7, 5, 2, 8, 9, 1, 5, 0, 6, 9, 9, 1, 0, 3, 7, 0, 9, 9, 6, 5, 5, 4, 3, 2, 5, 2, 3, 9, 9, 5, 5, 8, 1, 9, 8, 4, 7, 5, 9, 5, 9, 2, 6, 2, 9, 9, 7, 0, 2
OFFSET
0,2
COMMENTS
Decimal expansion of the shape of a lesser 2e-contraction rectangle.
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
Suppose that ABCD is given and that the shape r=[ABCD] exceeds 2. The "r-contraction rectangles" of ABCD are here introduced as the rectangles AEFD and EBCF for which [AEFD]=[EBCF] and |AE|<>|EB|. That is, ABCD has the prescribed shape r, and AEFD and EBCF are mutually similar without being congruent. It is easy to prove that [AEFD]=(1/2)(r-sqrt(-4+r^2)) or [AEFD]=(1/2)(r+sqrt(-4+r^2)); in the former case, we call AEFD the "lesser r-contraction rectangle", and the latter, the "greater r-contraction rectangle".
Both r-contraction rectangles match the continued fraction of [AEFD] in the following way. Write the continued fraction as [a(1),a(2),a(3),...]. Then, in the manner in which the continued fraction [1,1,1,...] matches the step-by-step removal of single squares from a golden triangle (as well as the manner in which the continued fraction [2,2,2,...] matches the step-by-step removal of 2 squares at a time from a silver triangle, etc.), remove a(1) squares at step 1, then remove a(2) squares at step 2, and so on, obtaining in the limit a partition of AEFD as an infinite set of squares.
For (related) r-extension rectangles, see A188640.
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.
EXAMPLE
0.190623604147330614259428256541555268663022202.. = 1/A188739 with continued fraction 0, 5, 4, 15, 6, 1, 13, 2, 1, 1, 21, 3, 2, 16, 1, 4, 1, 1, 157,...
MAPLE
evalf(exp(1)-sqrt(exp(2)-1), 140); # Muniru A Asiru, Nov 01 2018
MATHEMATICA
r = 2 E; t = (r - (-4 + r^2)^(1/2))/2; FullSimplify[t]
N[t, 130]
RealDigits[N[t, 130]][[1]]
ContinuedFraction[t, 120]
PROG
(PARI) default(realprecision, 100); exp(1) - sqrt(exp(2)-1) \\ G. C. Greubel, Nov 01 2018
(Magma) SetDefaultRealField(RealField(100)); Exp(1) - Sqrt(Exp(2)-1); // G. C. Greubel, Nov 01 2018
CROSSREFS
Cf. A001113, A188739 (inverse), A188627 (continued fraction), A188640.
Sequence in context: A345158 A019740 A372270 * A199789 A019874 A197520
KEYWORD
nonn,cons
AUTHOR
Clark Kimberling, Apr 11 2011
STATUS
approved