Displaying 41-50 of 659 results found.
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Beatty sequence for e: a(n) = floor(n*e).
+10
32
0, 2, 5, 8, 10, 13, 16, 19, 21, 24, 27, 29, 32, 35, 38, 40, 43, 46, 48, 51, 54, 57, 59, 62, 65, 67, 70, 73, 76, 78, 81, 84, 86, 89, 92, 95, 97, 100, 103, 106, 108, 111, 114, 116, 119, 122, 125, 127, 130, 133, 135, 138, 141, 144, 146, 149, 152, 154, 157, 160
FORMULA
a(n)/n converges to e because |a(n)/n-e|=|a(n)-n*e|/n < 1/n. - Hieronymus Fischer, Jan 22 2006
MATHEMATICA
Table[ Floor[n*E], {n, 1, 61}]
PROG
(Haskell)
a022843 n = a022843_list !! n
a022843_list = map (floor . (* e) . fromIntegral) [0..] where e = exp 1
(PARI) for (n=0, 100, print1(floor(n*exp(1)), ", ")) \\ Indranil Ghosh, Mar 21 2017 (Python)
import math
from mpmath import mp, e
mp.dps = 100
print([int(math.floor(n*e)) for n in range(51)]) # Indranil Ghosh, Mar 21 2017 (Magma) [Floor(n*Exp(1)): n in [0..60]]; // G. C. Greubel, Sep 28 2018
Decimal expansion of e^(1/e).
+10
32
1, 4, 4, 4, 6, 6, 7, 8, 6, 1, 0, 0, 9, 7, 6, 6, 1, 3, 3, 6, 5, 8, 3, 3, 9, 1, 0, 8, 5, 9, 6, 4, 3, 0, 2, 2, 3, 0, 5, 8, 5, 9, 5, 4, 5, 3, 2, 4, 2, 2, 5, 3, 1, 6, 5, 8, 2, 0, 5, 2, 2, 6, 6, 4, 3, 0, 3, 8, 5, 4, 9, 3, 7, 7, 1, 8, 6, 1, 4, 5, 0, 5, 5, 7, 3, 5, 8, 2, 9, 2, 3, 0, 4, 7, 0, 9, 8, 8, 5, 1, 1, 4, 2, 9, 5
COMMENTS
e^(1/e) = 1/((1/e)^(1/e)) (reciprocal of A072364). Let w(n+1)=A^w(n); then w(n) converges if and only if (1/e)^e <= A <= e^(1/e) (see the comments in A073230) for initial value w(1)=A. If A=e^(1/e) then lim_{n->infinity} w(n) = e. - Benoit Cloitre, Aug 06 2002; corrected by Robert FERREOL, Jun 12 2015 x^(1/x) is maximum for x = e and the maximum value is e^(1/e). This gives an interesting and direct proof that 2 < e < 4 as 2^(1/2) < e^(1/e) > 4^(1/4) while 2^(1/2) = 4^(1/4). - Amarnath Murthy, Nov 26 2002 Value of the unique base b > 0 for which the exponential curve y=b^x and its inverse y=log_b(x) kiss each other; the kissing point is (e,e). - Stanislav Sykora, May 25 2015 Actually, there is another base with such property, b=(1/e)^e with kiss point (1/e,1/e). - Yuval Paz, Dec 29 2018 The problem of finding the maximum of f(x) = x^(1/x) was posed and solved by the Swiss mathematician Jakob Steiner (1796-1863) in 1850. - Amiram Eldar, Jun 17 2021
REFERENCES
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 35.
LINKS
Ovidiu Furdui, Problem 11982, The American Mathematical Monthly, Vol. 124, No. 5 (2017), p. 465; A Limit of a Power of a Sum, Solution to Problem 11982 by Roberto Tauraso, ibid., Vol. 126, No. 2 (2019), p. 187.
FORMULA
Equals 1 + Integral_{x = 1/e..1} (1 + log(x))/x^x dx = 1 - Integral_{x = 0..1/e} (1 + log(x))/x^x dx. - Peter Bala, Oct 30 2019 Equals lim_{x->oo} (Sum_{n>=1} (x/n)^n)^(1/x) (Furdui, 2017). - Amiram Eldar, Mar 26 2022
EXAMPLE
1.44466786100976613365833910859...
MATHEMATICA
RealDigits[ E^(1/E), 10, 110] [[1]]
Decimal expansion of exp(-gamma).
+10
31
5, 6, 1, 4, 5, 9, 4, 8, 3, 5, 6, 6, 8, 8, 5, 1, 6, 9, 8, 2, 4, 1, 4, 3, 2, 1, 4, 7, 9, 0, 8, 8, 0, 7, 8, 6, 7, 6, 5, 7, 1, 0, 3, 8, 6, 9, 2, 5, 1, 5, 3, 1, 6, 8, 1, 5, 4, 1, 5, 9, 0, 7, 6, 0, 4, 5, 0, 8, 7, 9, 6, 7, 0, 7, 4, 2, 8, 5, 6, 3, 7, 1, 3, 2, 8, 7, 1, 1, 5, 8, 9, 3, 4, 2, 1, 4, 3, 5, 8, 7, 6, 7, 3, 1
COMMENTS
By Mertens's third theorem, lim_{k->oo} (H_{k-1}*Product_{prime p<=k} (1-1/p)) = exp(-gamma), where H_n is the n-th harmonic number. Let F(x) = lim_{n->oo} ((Sum_{k<=n} 1/k^x)*(Product_{prime p<=n} (1-1/p^x))) for real x in the interval 0 < x < 1. Consider the function F(s) of the complex variable s, but without the analytic continuation of the zeta function, in the critical strip 0 < Re(s) < 1. - Thomas Ordowski, Jan 26 2023
REFERENCES
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Sections 1.5 p. 29, 2.7 p. 117 and 5.4 p. 285.
LINKS
Steven R. Finch, Mathematical Constants II, Encyclopedia of Mathematics and Its Applications, Cambridge University Press, Cambridge, 2018, p. 202.
FORMULA
Equals lim_{k->oo} log(k)*Product_{prime p<=k} (1-1/p). - Amiram Eldar, Jul 09 2020 Equals Product_{k>=1} (1+1/k)*exp(-1/k). - Amiram Eldar, Mar 20 2022
EXAMPLE
0.56145948356688516982414321479088078676571...
PROG
(PARI) default(realprecision, 100); exp(-Euler) \\ G. C. Greubel, Aug 28 2018 (Magma) R:= RealField(100); Exp(-EulerGamma(R)); // G. C. Greubel, Aug 28 2018
CROSSREFS
Cf. A000010, A000142, A001113, A001620, A002852, A007838, A073004, A079650, A322364, A322365, A322380, A322381.
Decimal expansion of e/Pi.
+10
27
8, 6, 5, 2, 5, 5, 9, 7, 9, 4, 3, 2, 2, 6, 5, 0, 8, 7, 2, 1, 7, 7, 7, 4, 7, 8, 9, 6, 4, 6, 0, 8, 9, 6, 1, 7, 4, 2, 8, 7, 4, 4, 6, 2, 3, 9, 0, 8, 5, 1, 5, 5, 3, 9, 4, 5, 4, 3, 3, 0, 2, 8, 8, 9, 4, 8, 0, 4, 5, 0, 4, 4, 5, 7, 0, 6, 7, 7, 0, 5, 8, 6, 3, 1, 9, 2, 4, 6, 6, 2, 5, 1, 6, 1, 8, 4, 5, 1, 7, 2, 8, 6, 5, 8, 2
EXAMPLE
0.86525597943226508721777478964608961742874...
PROG
(PARI) { default(realprecision, 20080); x=10*exp(1)/Pi; for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b061360.txt", n, " ", d)); } \\ Harry J. Smith, Jul 21 2009
Decimal expansion of e + Pi.
+10
26
5, 8, 5, 9, 8, 7, 4, 4, 8, 2, 0, 4, 8, 8, 3, 8, 4, 7, 3, 8, 2, 2, 9, 3, 0, 8, 5, 4, 6, 3, 2, 1, 6, 5, 3, 8, 1, 9, 5, 4, 4, 1, 6, 4, 9, 3, 0, 7, 5, 0, 6, 5, 3, 9, 5, 9, 4, 1, 9, 1, 2, 2, 2, 0, 0, 3, 1, 8, 9, 3, 0, 3, 6, 6, 3, 9, 7, 5, 6, 5, 9, 3, 1, 9, 9, 4, 1, 7, 0, 0, 3, 8, 6, 7, 2, 8, 3, 4, 9, 5, 4, 0, 9, 6, 1
COMMENTS
It is not presently known if this number is rational or irrational.
EXAMPLE
5.859874482048838473822930854632165381954416493075065395941912220031893...
MAPLE
Digits := 200: with(numtheory): it := evalf((Pi+exp(1))/10, 200): for i from 1 to 20 0 do printf(`%d, `, floor(10*it)): it := 10*it-floor(10*it): od:
PROG
(PARI) { default(realprecision, 20080); x=Pi+exp(1); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b059742.txt", n, " ", d)); } \\ Harry J. Smith, May 31 2009
Expansion of e in base 2.
+10
24
1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1
COMMENTS
The first 19 digits are a reverse copy of the 7th through 25th digits of the binary expansion of Pi ( A004601). - Dan Graham, May 03 2015
CROSSREFS
Expansion of e in base b: this sequence (b=2), A004594 (b=3), A004595 (b=4), A004596 (b=5), A004597 (b=6), A004598 (b=7), A004599 (b=8), A004600 (b=9), A001113 (b=10), A170873 (b=16). - Jason Kimberley, Dec 05 2012
Decimal expansion of -exp(1)*Ei(-1), also called Gompertz's constant, or the Euler-Gompertz constant.
+10
23
5, 9, 6, 3, 4, 7, 3, 6, 2, 3, 2, 3, 1, 9, 4, 0, 7, 4, 3, 4, 1, 0, 7, 8, 4, 9, 9, 3, 6, 9, 2, 7, 9, 3, 7, 6, 0, 7, 4, 1, 7, 7, 8, 6, 0, 1, 5, 2, 5, 4, 8, 7, 8, 1, 5, 7, 3, 4, 8, 4, 9, 1, 0, 4, 8, 2, 3, 2, 7, 2, 1, 9, 1, 1, 4, 8, 7, 4, 4, 1, 7, 4, 7, 0, 4, 3, 0, 4, 9, 7, 0, 9, 3, 6, 1, 2, 7, 6, 0, 3, 4, 4, 2, 3, 7
COMMENTS
0! - 1! + 2! - 3! + 4! - 5! + ... = (Borel) Sum_{n>=0} (-y)^n n! = KummerU(1,1,1/y)/y.
Decimal expansion of phi(1) where phi(x) = Integral_{t>=0} e^-t/(x+t) dt. - Benoit Cloitre, Apr 11 2003 The divergent series g(x=1,m) = 1^m*1! - 2^m*2! + 3^m*3! - 4^m*4! + ..., m => -1, is intimately related to Gompertz's constant. We discovered that g(x=1,m) = (-1)^m * ( A040027(m) - A000110(m+1) * A073003) with A000110 the Bell numbers and A040027 a sequence that was published by Gould, see for more information A163940. - Johannes W. Meijer, Oct 16 2009 Named by Le Lionnais (1983) after the English self-educated mathematician and actuary Benjamin Gompertz (1779 - 1865). It was named the Euler-Gompertz constant by Finch (2003). Lagarias (2013) noted that he has not located this constant in Gompertz's writings. - Amiram Eldar, Aug 15 2020
REFERENCES
Bruce C. Berndt, Ramanujan's notebooks Part II, Springer, p. 171
Bruce C. Berndt, Ramanujan's notebooks Part I, Springer, p. 144-145.
S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 424-425.
Francois Le Lionnais, Les nombres remarquables, Paris: Hermann, 1983. See p. 29.
H. S. Wall, Analytic Theory of Continued Fractions, Van Nostrand, New York, 1948, p. 356.
FORMULA
phi(1) = e*(Sum_{k>=1} (-1)^(k-1)/(k*k!) - Gamma) = 0.596347362323194... where Gamma is the Euler constant.
G = 0.596347... = 1/(1+1/(1+1/(1+2/(1+2/(1+3/(1+3/(1+4/(1+4/(1+5/(1+5/(1+6/(... - Philippe Deléham, Aug 14 2005 Stieltjes found the continued fraction representation G = 1/(2 - 1^2/(4 - 2^2/(6 - 3^2/(8 - ...)))). See [Wall, Chapter 18, (92.7) with a = 1]. The sequence of convergents to the continued fraction begins [1/2, 4/7, 20/34, 124/209, ...]. The numerators are in A002793 and the denominators in A002720. Also, 1 - G has the continued fraction representation 1/(3 - 2/(5 - 6/(7 - ... -n*(n+1)/((2*n+3) - ...)))) with convergents beginning [1/3, 5/13, 29/73, 201/501, ...]. The numerators are in A201203 (unsigned) and the denominators are in A000262. (End)
G = f(1) with f solution to the o.d.e. x^2*f'(x) + (x+1)*f(x)=1 such that f(0)=1. - Jean-François Alcover, May 28 2013 Equals Integral_{x=0..1} 1/(1-log(x)) dx.
Equals Integral_{x=1..oo} exp(1-x)/x dx.
Equals Integral_{x=0..oo} exp(-x)*log(x+1) dx.
Equals Integral_{x=0..oo} exp(-x)/(x+1) dx. (End)
Equals Integral_{x=0..1} LambertW(e/x)-1 dx.
Equals Integral_{x=0..1} 1+1/LambertW(-1,-x/e) dx. (End)
Equals Sum_{n >= 1} 1/(n*L(n, -1)*L(n-1, -1)), where L(n, x) denotes the n-th Laguerre polynomial. This is the case x = 1 of the identity Integral_{t >= 0} exp(-t)/(x + t) dt = Sum_{n >= 1} 1/(n*L(n, -x)*L(n-1, -x)) valid for Re(x) > 0. - Peter Bala, Mar 21 2024 Equals lim_{n -> oo} Sum_{k >= 0} (n/(n + 1))^k/(n + k). Cf. A099285. - Peter Bala, Jun 18 2024
EXAMPLE
0.59634736232319407434107849936927937607417786015254878157348491...
With n := 10^5, Sum_{k >= 0} (n/(n + 1))^k/(n + k) = 0.5963(51...). - Peter Bala, Jun 19 2024
MATHEMATICA
RealDigits[N[-Exp[1]*ExpIntegralEi[-1], 105]][[1]]
(* Second program: *)
G = 1/Fold[Function[2*#2 - #2^2/#1], 2, Reverse[Range[10^4]]] // N[#, 105]&; RealDigits[G] // First (* Jean-François Alcover, Sep 19 2014 *)
PROG
(Magma) SetDefaultRealField(RealField(100)); ExponentialIntegralE1(1)*Exp(1); // G. C. Greubel, Dec 04 2018 (Sage) numerical_approx(exp_integral_e(1, 1)*exp(1), digits=100) # G. C. Greubel, Dec 04 2018
CROSSREFS
Cf. A000522 (arrangements), A001620, A000262, A002720, A002793, A058006 (alternating factorial sums), A091725, A099285, A153229, A201203, A245780, A283743 (Ei(1)/e), A321942, A369883.
0, 2, 12, 72, 480, 3600, 30240, 282240, 2903040, 32659200, 399168000, 5269017600, 74724249600, 1133317785600, 18307441152000, 313841848320000, 5690998849536000, 108840352997376000, 2189611807358976000, 46225138155356160000, 1021818843434188800000
COMMENTS
For n > 0, a(n) = number of permutations of length n+1 that have 2 predetermined elements nonadjacent; e.g., for n=2, the permutations with, say, 1 and 2 nonadjacent are 132 and 231, therefore a(2)=2. - Jon Perry, Jun 08 2003 Number of multiplications performed when computing the determinant of an n X n matrix by definition. - Mats Granvik, Sep 12 2008 Sum of the length of all cycles (excluding fixed points) in all permutations of [n]. - Olivier Gérard, Oct 23 2012 Number of permutations of n distinct objects (ABC...) 1 (one) times >>("-", A, AB, ABC, ABCD, ABCDE, ..., ABCDEFGHIJK, infinity) and one after the other to resemble motif: A (1) AB (1-1), AAB (2-1), AAAB (3-1), AAAAB (4-1), AAAAAB (5-1), AAAAAAB (6-1), AAAAAAAB (7-1), AAAAAAAAB (8-1) etc.,>> "1(one) fixed point". Example:motif: AAAB (or BBBA) 12 * one (1) fixed point etc. Let: AAAB ................ 'A'BCD 1. 'A'BDC 2. 'A'CBD 3. ACDB 'A'DBC 4. 'A'DCB B'A'CD 5. B'A'DC 6. BCAD 7. BCDA BD'A'C 8. BDCA C'A'BD 9. C'A'DB CB'A'D 10. CBDA CDAB CDBA D'A'BC 11. DACB DB'A'C 12. DBCA DCAB DCBA. - Zerinvary Lajos, Nov 27 2009 (does anybody understand what this is supposed to say? - Joerg Arndt, Jan 10 2015) a(n) is the number of ways to arrange n books on two bookshelves so that each shelf receives at least one book. - Geoffrey Critzer, Feb 21 2010 a(n) = number whose factorial base representation ( A007623) begins with digit {n-1} and is followed by n-1 zeros. Viewed in that base, this sequence looks like this: 0, 10, 200, 3000, 40000, 500000, 6000000, 70000000, 800000000, 9000000000, A0000000000, B00000000000, ... (where "digits" A and B stand for placeholder values 10 and 11 respectively). - Antti Karttunen, May 07 2015
MAPLE
G(x):=x^2/(1-x)^2: f[0]:=G(x): for n from 1 to 19 do f[n]:=diff(f[n-1], x) od: x:=0: seq(f[n], n=1..19); # Zerinvary Lajos, Apr 01 2009
PROG
(PARI) { f=1; for (n=1, 100, f*=n; write("b062119.txt", n, " ", f*(n - 1)) ) } \\ Harry J. Smith, Aug 02 2009 (Haskell)
CROSSREFS
Column 2 of A257503 (apart from initial zero. Equally, row 2 of A257505). Cf. A001286 (same sequence divided by 2). Cf. sequences with formula (n + k)*n! listed in A282466.
a(n) = (n-1)*(n+1)!/6.
(Formerly M4551)
+10
21
0, 1, 8, 60, 480, 4200, 40320, 423360, 4838400, 59875200, 798336000, 11416204800, 174356582400, 2833294464000, 48819843072000, 889218570240000, 17072996548608000, 344661117825024000, 7298706024529920000, 161787983543746560000
COMMENTS
Coefficients of Gandhi polynomials.
a(n) = Sum_{pi in Symm(n)} Sum_{i=1..n} max(pi(i)-i,0), i.e., the total positive displacement of all letters in all permutations on n letters. - Franklin T. Adams-Watters, Oct 25 2006 a(n) is also the sum of the excedances of all permutations of [n]. An excedance of a permutation p of [n] is an i (1 <= i <= n-1) such that p(i) > i. Proof: i is an excedance if p(i) = i+1, i+2, ..., n (n-i possibilities), with the remaining values of p forming any permutation of [n]\{p(i)} in the positions [n]\{i} ((n-1)! possibilities). Summation of i(n-i)(n-1)! over i from 1 to n-1 completes the proof. Example: a(3)=8 because the permutations 123, 132, 213, 231, 312, 321 have excedances NONE, {2}, {1}, {1,2}, {1}, {1}, respectively. - Emeric Deutsch, Oct 26 2008 a(n) is also the number of doubledescents in all permutations of {1,2,...,n-1}. We say that i is a doubledescent of a permutation p if p(i) > p(i+1) > p(i+2). Example: a(3)=8 because each of the permutations 1432, 4312, 4213, 2431, 3214, 3421 has one doubledescent, the permutation 4321 has two doubledescents and the remaining 17 permutations of {1,2,3,4} have no doubledescents. - Emeric Deutsch, Jul 26 2009 Half of sum of abs(p(i+1) - p(i)) over all permutations on n, e.g., 42531 = 2 + 3 + 2 + 2 = 9, and the total over all permutations on {1,2,3,4,5} is 960. - Jon Perry, May 24 2013 a(n) gives the number of non-occupied corners in tree-like tableaux of size n+1 (see Gao et al. link). - Michel Marcus, Nov 18 2015 a(n) is the number of sequences of n+2 balls colored with at most n colors such that exactly three balls are the same color as some other ball in the sequence. - Jeremy Dover, Sep 26 2017 a(n) is the number of triangles (3-cycles) in the (n+1)-alternating group graph. - Eric W. Weisstein, Jun 09 2019
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Alice L. L. Gao, Emily X. L. Gao, and Brian Y. Sun, Zubieta's Conjecture on the Enumeration of Corners in Tree-like Tableaux, arXiv:1511.05434 [math.CO], 2015. The second version of this paper has a different title and different authors: A. L. L. Gao, E. X. L. Gao, P. Laborde-Zubieta, and B. Y. Sun, Enumeration of Corners in Tree-like Tableaux and a Conjectural (a,b)-analogue, arXiv preprint arXiv:1511.05434v2, 2015.
FORMULA
a(n) = Sum_{m=0..n} Sum_{k=-1..n} Sum_{j=1..n} n!/6, n >= 0. - Zerinvary Lajos, May 11 2007 If we define f(n,i,x) = Sum_{k=i..n} (Sum_{j=i..k} binomial(k,j)*Stirling1(n,k)*Stirling2(j,i)*x^(k-j)) then a(n+1) = (-1)^(n-1)*f(n,1,-4), (n >= 1). - Milan Janjic, Mar 01 2009 E.g.f.: (-1+3*x)/(3!*(1-x)^3), a(0) = -1/3!. Such e.g.f. computations resulted from e-mail exchange with Gary Detlefs. - Wolfdieter Lang, May 27 2010 a(n) = ((n+3)!/2) * Sum_{j=i..k} (k+1)!/(k+3)!, with offset 0. - Gary Detlefs, Aug 05 2010 a(n) = (n+2)!*Sum_{k=1..n-1} 1/((2*k+4)*(k+3)). - Gary Detlefs, Oct 09 2011 a(n) = (n+2)!*(1 + 3*(H(n+1) - H(n+2)))/6, where H(n) is the n-th harmonic number. - Gary Detlefs, Oct 09 2011 Sum_{n>=2} 1/a(n) = 3*(Ei(1) - gamma) - 6*e + 27/2, where Ei(1) = A091725, gamma = A001620, and e = A001113. Sum_{n>=2} (-1)^n/a(n) = 3*(gamma - Ei(-1)) - 3/2, where Ei(-1) = - A099285. (End)
MAPLE
[ seq((n-1)*(n+1)!/6, n=1..40) ];
a:=n->sum(sum(sum(n!/6, j=1..n), k=-1..n), m=0..n): seq(a(n), n=0..19); # Zerinvary Lajos, May 11 2007 seq(sum(mul(j, j=3..n), k=3..n)/3, n=2..21); # Zerinvary Lajos, Jun 01 2007 restart: G(x):=x^3/(1-x)^2: f[0]:=G(x): for n from 1 to 21 do f[n]:=diff(f[n-1], x) od: x:=0: seq(f[n]/3!, n=2..21); # Zerinvary Lajos, Apr 01 2009
MATHEMATICA
Range[0, 20]! CoefficientList[Series[x/(1 - x)^4, {x, 0, 20}], x] (* Eric W. Weisstein, Jun 09 2019 *)
EXTENSIONS
Better definition from Robert Newstedt
Decimal expansion of e-sqrt(e^2-1).
+10
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
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.
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,...
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
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