cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 25 results. Next

A014176 Decimal expansion of the silver mean, 1+sqrt(2).

Original entry on oeis.org

2, 4, 1, 4, 2, 1, 3, 5, 6, 2, 3, 7, 3, 0, 9, 5, 0, 4, 8, 8, 0, 1, 6, 8, 8, 7, 2, 4, 2, 0, 9, 6, 9, 8, 0, 7, 8, 5, 6, 9, 6, 7, 1, 8, 7, 5, 3, 7, 6, 9, 4, 8, 0, 7, 3, 1, 7, 6, 6, 7, 9, 7, 3, 7, 9, 9, 0, 7, 3, 2, 4, 7, 8, 4, 6, 2, 1, 0, 7, 0, 3, 8, 8, 5, 0, 3, 8, 7, 5, 3, 4, 3, 2, 7, 6, 4, 1, 5, 7
Offset: 1

Views

Author

Keywords

Comments

From Hieronymus Fischer, Jan 02 2009: (Start)
Set c:=1+sqrt(2). Then the fractional part of c^n equals 1/c^n, if n odd. For even n, the fractional part of c^n is equal to 1-(1/c^n).
c:=1+sqrt(2) satisfies c-c^(-1)=floor(c)=2, hence c^n + (-c)^(-n) = round(c^n) for n>0, which follows from the general formula of A001622.
1/c = sqrt(2)-1.
See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).
Other examples of constants x satisfying the relation x-x^(-1)=floor(x) include A001622 (the golden ratio: where floor(x)=1) and A098316 (the "bronze" ratio: where floor(x)=3). (End)
In terms of continued fractions the constant c can be described by c=[2;2,2,2,...]. - Hieronymus Fischer, Oct 20 2010
Side length of smallest square containing five circles of diameter 1. - Charles R Greathouse IV, Apr 05 2011
Largest radius of four circles tangent to a circle of radius 1. - Charles R Greathouse IV, Jan 14 2013
An analog of Fermat theorem: for prime p, round(c^p) == 2 (mod p). - Vladimir Shevelev, Mar 02 2013
n*(1+sqrt(2)) is the perimeter of a 45-45-90 triangle with hypotenuse n. - Wesley Ivan Hurt, Apr 09 2016
This algebraic integer of degree 2, with minimal polynomial x^2 - 2*x - 1, is also the length ratio diagonal/side of the second largest diagonal in the regular octagon (not counting the side). The other two diagonal/side ratios are A179260 and A121601. - Wolfdieter Lang, Oct 28 2020
c^n = A001333(n) + A000129(n) * sqrt(2). - Gary W. Adamson, Apr 26 2023
c^n = c * A000129(n) + A000129(n-1), where c = 1 + sqrt(2). - Gary W. Adamson, Aug 30 2023

Examples

			2.414213562373095...
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.

Crossrefs

Apart from initial digit the same as A002193.
See A098316 for [3;3,3,...]; A098317 for [4;4,4,...]; A098318 for [5;5,5,...]. - Hieronymus Fischer, Oct 20 2010

Programs

  • Maple
    Digits:=100: evalf(1+sqrt(2)); # Wesley Ivan Hurt, Apr 09 2016
  • Mathematica
    RealDigits[1 + Sqrt@ 2, 10, 111] (* Or *)
    RealDigits[Exp@ ArcSinh@ 1, 10, 111][[1]] (* Robert G. Wilson v, Aug 17 2011 *)
    Circs[n_] := With[{r = Sin[Pi/n]/(1 - Sin[Pi/n])}, Graphics[Append[
      Table[Circle[(r + 1) {Sin[2 Pi k/n], Cos[2 Pi k/n]}, r], {k, n}],   {Blue, Circle[{0, 0}, 1]}]]] Circs[4] (* Charles R Greathouse IV, Jan 14 2013 *)
  • PARI
    1+sqrt(2) \\ Charles R Greathouse IV, Jan 14 2013

Formula

Conjecture: 1+sqrt(2) = lim_{n->oo} A179807(n+1)/A179807(n).
Equals cot(Pi/8) = tan(Pi*3/8). - Bruno Berselli, Dec 13 2012, and M. F. Hasler, Jul 08 2016
Silver mean = 2 + Sum_{n>=0} (-1)^n/(P(n-1)*P(n)), where P(n) is the n-th Pell number (A000129). - Vladimir Shevelev, Feb 22 2013
Equals exp(arcsinh(1)) which is exp(A091648). - Stanislav Sykora, Nov 01 2013
Limit_{n->oo} exp(asinh(cos(Pi/n))) = sqrt(2) + 1. - Geoffrey Caveney, Apr 23 2014
exp(asinh(cos(Pi/2 - log(sqrt(2)+1)*i))) = exp(asinh(sin(log(sqrt(2)+1)*i))) = i. - Geoffrey Caveney, Apr 23 2014
Equals Product_{k>=1} A047621(k) / A047522(k) = (3/1) * (5/7) * (11/9) * (13/15) * (19/17) * (21/23) * ... . - Dimitris Valianatos, Mar 27 2019
From Wolfdieter Lang, Nov 10 2023:(Start)
Equals lim_{n->oo} A000129(n+1)/A000129(n) (see A000129, Pell).
Equals lim_{n->oo} S(n+1, 2*sqrt(2))/S(n, 2*sqrt(2)), with the Chebyshev S(n,x) polynomial (see A049310). (End)
From Peter Bala, Mar 24 2024: (Start)
An infinite family of continued fraction expansions for this constant can be obtained from Berndt, Entry 25, by setting n = 1/2 and x = 8*k + 6 for k >= 0.
For example, taking k = 0 and k = 1 yields
sqrt(2) + 1 = 15/(6 + (1*3)/(12 + (5*7)/(12 + (9*11)/(12 + (13*15)/(12 + ... + (4*n + 1)*(4*n + 3)/(12 + ... )))))) and
sqrt(2) + 1 = (715/21) * 1/(14 + (1*3)/(28 + (5*7)/(28 + (9*11)/(28 + (13*15)/(28 + ... + (4*n + 1)*(4*n + 3)/(28 + ... )))))). (End)

A035185 Number of divisors of n == 1 or 7 (mod 8) minus number of divisors of n == 3 or 5 (mod 8).

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 2, 1, 1, 0, 0, 0, 0, 2, 0, 1, 2, 1, 0, 0, 0, 0, 2, 0, 1, 0, 0, 2, 0, 0, 2, 1, 0, 2, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 2, 2, 0, 3, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 2, 1, 0, 0, 0, 2, 0, 0, 2, 1, 2, 0, 0, 0, 0, 0, 2, 0, 1, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 2, 0, 2, 0, 0, 2, 3, 0, 1, 0, 0, 2, 0, 0
Offset: 1

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Author

Keywords

Comments

Coefficients in expansion of Dirichlet series Product_p (1-(Kronecker(m,p)+1)*p^(-s) + Kronecker(m,p)*p^(-2s))^(-1) for m = 2.
Let zetaQ(sqrt(2))(s) = Sum (1/(Z(sqrt(2)):A)^s), a Dedekind zeta function, where A runs through the nonzero ideals of Z(sqrt(2)) and where (Z(sqrt(2)):A) is the norm of A; then zetaQ(sqrt(2))(s) = Sum_{n>=1}, a(n)/n^s); Sum{k=1..n} a(k) is asymptotic to c*n where c = log(1 + sqrt(2))/sqrt(2). - Benoit Cloitre, Jan 01 2003
Inverse Moebius transform of A091337.
a(n) is the number of solutions to the equation n = x^2 - 2*y^2 in integers where -x < 2*y <= x. [Uspensky and Heaslet] - Michael Somos, Feb 17 2020
Coefficients of Dedekind zeta function for the quadratic number field of discriminant 8. See A002324 for formula and Maple code. - N. J. A. Sloane, Mar 22 2022

Examples

			G.f. = x + x^2 + x^4 + 2*x^7 + x^8 + x^9 + 2*x^14 + x^16 + 2*x^17 + x^18 + ...
a(7) = 2 because 7 = 3^2 - 2*(+1)^2 = 3^2 - 2*(-1)^2. - _Michael Somos_, Feb 17 2020
		

References

  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 368.

Crossrefs

Moebius transform gives A091337.
Dedekind zeta functions for imaginary quadratic number fields of discriminants -3, -4, -7, -8, -11, -15, -19, -20 are A002324, A002654, A035182, A002325, A035179, A035175, A035171, A035170, respectively.
Dedekind zeta functions for real quadratic number fields of discriminants 5, 8, 12, 13, 17, 21, 24, 28, 29, 33, 37, 40 are A035187, A035185, A035194, A035195, A035199, A035203, A035188, A035210, A035211, A035215, A035219, A035192, respectively.

Programs

  • Mathematica
    a[ n_] := If[ n < 1, 0, DivisorSum[ n, KroneckerSymbol[ 2, #] &]]; (* Michael Somos, Jul 06 2015 *)
    a[ n_] := SeriesCoefficient[ Sum[ x^k (1 - x^(2 k)) / (1 + x^(4 k)), {k, n}], {x, 0, n}]; (* Michael Somos, Jul 06 2015 *)
    a[ n_] := If[ n < 1, 0, Times @@ (Which[ # <= 2, 1, Mod[#, 8] > 1 && Mod[#, 8] < 7, 1 - Mod[#2, 2], True, #2 + 1] & @@@ FactorInteger @ n)]; (* Michael Somos, Jul 06 2015 *)
  • PARI
    {a(n) = if( n<1, 0, sumdiv(n, d, kronecker(2, d)))};
    
  • PARI
    {a(n) = if( n<1, 0, direuler(p=2, n, 1 / ((1 - X) * (1 - kronecker(2, p)*X)))[n])};
    
  • PARI
    {a(n) = my(A, p, e); if( n<1, 0, A = factor(n); prod(k=1, matsize(A)[1], [p, e] = A[k, ]; if(p==2, 1, p%8>1 && p%8<7, !(e%2), e+1)))}; \\ Michael Somos, Aug 17 2006
    
  • PARI
    {a(n) = if( n<1, 0, polcoeff( sum(k=1, n, x^k * (1 - x^(2*k)) / (1 + x^(4*k)), x * O(x^n)), n))}; \\ Michael Somos, Jul 06 2015

Formula

G.f.: Sum_{k>0} x^k * (1 - x^(2*k)) / (1 + x^(4*k)).
-(-1)^(n*(n-1)/2)*a(n) = Sum_{n >= 1} (-1)^n * q^(n*(n+1)/2)*(1-q)*(1-q^2)*...*(1-q^(n-1))/ ((1+q)*(1+q^2)*...*(1+q^n)). - Jeremy Lovejoy, Jun 12 2009
a(n) = (-1)^floor(n/2) * A259829(n). - Michael Somos, Jul 06 2015
a(n) is multiplicative with a(2^e) = 1, a(p^e) = (1 + (-1)^e) / 2 if p == 3, 5 (mod 8), a(p^e) = e + 1 if p == 1, 7 (mod 8). - Jianing Song, Sep 07 2018
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = log(sqrt(2)+1)/sqrt(2) = A091648/A002193 = 0.623225... . - Amiram Eldar, Oct 11 2022

A244920 Decimal expansion of 2*log(1+sqrt(2)), the integral over the square [0,1]x[0,1] of 1/sqrt(x^2+y^2) dx dy.

Original entry on oeis.org

1, 7, 6, 2, 7, 4, 7, 1, 7, 4, 0, 3, 9, 0, 8, 6, 0, 5, 0, 4, 6, 5, 2, 1, 8, 6, 4, 9, 9, 5, 9, 5, 8, 4, 6, 1, 8, 0, 5, 6, 3, 2, 0, 6, 5, 6, 5, 2, 3, 2, 7, 0, 8, 2, 1, 5, 0, 6, 5, 9, 1, 2, 1, 7, 3, 0, 6, 7, 5, 4, 3, 6, 8, 4, 4, 4, 0, 5, 2, 1, 7, 5, 6, 6, 7, 4, 1, 3, 7, 8, 3, 8, 2, 0, 5, 1, 2, 0, 8, 5, 7
Offset: 1

Views

Author

Jean-François Alcover, Jul 08 2014

Keywords

Comments

Number field regulator of the cyclotomic number field Q(zeta_8), where zeta_8 = sqrt(i), an eighth root of 1. - Alonso del Arte, Mar 11 2017

Examples

			1.7627471740390860504652186499595846180563206565232708215065912173...
		

Crossrefs

Equals twice A091648. - Michel Marcus, Mar 18 2017
Cf. A156035.

Programs

  • Mathematica
    RealDigits[2 * Log[1 + Sqrt[2]], 10, 101] // First
    RealDigits[NumberFieldRegulator[Sqrt[I]], 10, 100][[1]] (* Alonso del Arte, Mar 11 2017 *)
  • PARI
    2*asinh(1) \\ Michel Marcus, Mar 18 2017

Formula

Equals 2*arcsinh(1).
Equals Integral_{x>=1} 1/(x*(1+x)^(1/2)) dx. - Pointed out by Robert FERREOL.
Equals arccosh(3). - Vaclav Kotesovec, Dec 11 2016
Equals Integral_{x>=1} arcsinh(x)/x^2 dx. - Amiram Eldar, Jun 26 2021
Equals Integral_{x = 0..Pi/2} x/cos(x/2) dx. - Peter Bala, Aug 13 2024
Equals log(A156035). - Hugo Pfoertner, Aug 17 2024
Equals arcsinh(2*sqrt(2)). - Akiva Weinberger, Dec 03 2024
Equals Integral_{x=0..oo} erf(sqrt(x))/(x*e^x) dx. - Kritsada Moomuang, May 25 2025

A103710 Decimal expansion of the ratio of the length of the latus rectum arc of any parabola to its semi latus rectum: sqrt(2) + log(1 + sqrt(2)).

Original entry on oeis.org

2, 2, 9, 5, 5, 8, 7, 1, 4, 9, 3, 9, 2, 6, 3, 8, 0, 7, 4, 0, 3, 4, 2, 9, 8, 0, 4, 9, 1, 8, 9, 4, 9, 0, 3, 8, 7, 5, 9, 7, 8, 3, 2, 2, 0, 3, 6, 3, 8, 5, 8, 3, 4, 8, 3, 9, 2, 9, 9, 7, 5, 3, 4, 6, 6, 4, 4, 1, 0, 9, 6, 6, 2, 6, 8, 4, 1, 3, 3, 1, 2, 6, 6, 8, 4, 0, 9, 4, 4, 2, 6, 2, 3, 7, 8, 9, 7, 6, 1, 5, 5, 9, 1, 7, 5
Offset: 1

Views

Author

Sylvester Reese and Jonathan Sondow, Feb 13 2005

Keywords

Comments

The universal parabolic constant, equal to the ratio of the latus rectum arc of any parabola to its focal parameter. Like Pi, it is transcendental.
Just as all circles are similar, all parabolas are similar. Just as the ratio of a semicircle to its radius is always Pi, the ratio of the latus rectum arc of any parabola to its semi latus rectum is sqrt(2) + log(1 + sqrt(2)).
Note the remarkable similarity to sqrt(2) - log(1 + sqrt(2)), the universal equilateral hyperbolic constant A222362, which is a ratio of areas rather than of arc lengths. Lockhart (2012) says "the arc length integral for the parabola .. is intimately connected to the hyperbolic area integral ... I think it is surprising and wonderful that the length of one conic section is related to the area of another."
Is it a coincidence that the universal parabolic constant is equal to 6 times the expected distance A103712 from a randomly selected point in the unit square to its center? (Reese, 2004; Finch, 2012)

Examples

			2.29558714939263807403429804918949038759783220363858348392997534664...
		

References

  • H. Dörrie, 100 Great Problems of Elementary Mathematics, Dover, 1965, Problems 57 and 58.
  • P. Lockhart, Measurement, Harvard University Press, 2012, p. 369.
  • C. E. Love, Differential and Integral Calculus, 4th ed., Macmillan, 1950, pp. 286-288.
  • C. S. Ogilvy, Excursions in Geometry, Oxford Univ. Press, 1969, p. 84.
  • S. Reese, A universal parabolic constant, 2004, preprint.

Crossrefs

Programs

  • Mathematica
    RealDigits[ Sqrt[2] + Log[1 + Sqrt[2]], 10, 111][[1]] (* Robert G. Wilson v Feb 14 2005 *)
  • Maxima
    fpprec: 100$ ev(bfloat(sqrt(2) + log(1 + sqrt(2)))); /* Martin Ettl, Oct 17 2012 */
    
  • PARI
    sqrt(2)+log(1+sqrt(2)) \\ Charles R Greathouse IV, Mar 08 2013

Formula

Equals 2*Integral_{x = 0..1} sqrt(1 + x^2) dx. - Peter Bala, Feb 28 2019

A181048 Decimal expansion of (log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4*k+1).

Original entry on oeis.org

8, 6, 6, 9, 7, 2, 9, 8, 7, 3, 3, 9, 9, 1, 1, 0, 3, 7, 5, 7, 3, 9, 9, 5, 1, 6, 3, 8, 8, 2, 8, 7, 0, 7, 1, 3, 6, 5, 2, 1, 7, 5, 3, 6, 7, 3, 4, 5, 2, 4, 4, 9, 0, 4, 3, 3, 5, 0, 3, 1, 8, 3, 8, 9, 1, 7, 6, 3, 9, 3, 5, 1, 4, 1, 0, 9, 4, 1, 3, 2, 9, 0, 5, 5, 7, 5, 0, 4, 0, 3, 4, 6, 3, 4, 0, 8, 9, 6, 8, 7, 0, 5, 2, 1, 8
Offset: 0

Views

Author

Jonathan D. B. Hodgson, Oct 01 2010, Oct 06 2010

Keywords

Examples

			0.86697298733991103757399516388287071365217536734524490433....
At N = 100000 the truncated series 2*Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 1) = 1.7339(3)5974(5)7982(5)075(25)79(846)27(404)7... to 32 digits. The bracketed numbers show where this decimal expansion differs from that of 2*A181048. The numbers 1, 1, -3, -11, 57, 361 must be added to the bracketed numbers to give the correct decimal expansion to 32 digits: 2*( (log(1 + sqrt(2)) + Pi/2)/(2*sqrt(2)) ) = 1.7339(4)5974(6)7982(2)075(14)79(903)27(765)7.... - _Peter Bala_, Sep 23 2016
		

References

  • Jolley, Summation of Series, Dover (1961) eq 82 page 16.
  • Murray R. Spiegel, Seymour Lipschutz, John Liu. Mathematical Handbook of Formulas and Tables, 3rd Ed. Schaum's Outline Series. New York: McGraw-Hill (2009): p. 135, equation 21.17

Crossrefs

Programs

  • Mathematica
    RealDigits[(Pi Sqrt[2])/8 + (Sqrt[2] Log[1 + Sqrt[2]])/4, 10, 100][[1]] (* Alonso del Arte, Aug 11 2011 *)
  • PARI
    (log(1+sqrt(2))+Pi/2)/(2*sqrt(2)) \\ G. C. Greubel, Jul 05 2017
    
  • PARI
    (asinh(1)+Pi/2)/sqrt(8) \\ Charles R Greathouse IV, Jul 06 2017

Formula

Equals (A093954 + A091648/sqrt(2))/2.
Integral_{x = 0..1} 1/(1+x^4) = Sum_{k >= 0} (-1)^k/(4*k+1) = (log(1+sqrt(2)) + Pi/2)/(2*sqrt(2)).
1 - 1/5 + 1/9 - 1/13 + 1/17 - ... = (Pi*sqrt(2))/8 + (sqrt(2)*log(1 + sqrt(2)))/4 = (Pi + 2*log(1 + sqrt(2)))/(4 sqrt(2)). The first two are the formulas as given in Spiegel et al., the third is how Mathematica rewrites the infinite sum. - Alonso del Arte, Aug 11 2011
Let N be a positive integer divisible by 4. We have the asymptotic expansion 2*( (log(1 + sqrt(2)) + Pi/2)/(2*sqrt(2)) - Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 1) ) ~ 1/N + 1/N^2 - 3/N^3 - 11/N^4 + 57/N^5 + 361/N^6 - ..., where the sequence of coefficients [1, 1, -3, -11, 57, 361, ...] is A188458. This follows from Borwein et al., Lemma 2 with f(x) = 1/x and then set x = N/4 and h = 1/4. An example is given below. Cf. A181049. - Peter Bala, Sep 23 2016
Equals Sum_{n >= 0} 2^(n-1)*n!/(Product_{k = 0..n} 4*k + 1) = Sum_{n >= 0} 2^(n-1)*n!/A007696(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(4*k + 1)). - Peter Bala, Dec 01 2021
From Peter Bala, Oct 23 2023: (Start)
The slowly converging series representation Sum_{n >= 0} (-1)^n/(4*n + 1) for the constant can be accelerated to give the following faster converging series:
1/2 + 2*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5));
7/10 + 8*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9));
71/90 + 48*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13));
971/1170 + 384*Sum_{n >= 0} (-1)^n/((4*n + 1)(4*n + 5)*(4*n + 9)*(4*n + 13)*(4*n + 17)).
These results may be easily verified by taking the partial fraction expansions of the summands. The general result appears to be that for r >= 0, the constant equals
C(r) + (2^r)*r!*Sum_{n >= 0} (-1)^n/((4*n + 1)*(4*n + 5)*...*(4*n + 4*r + 1)), where C(r) is the rational number Sum_{k = 0..r-1} 2^(k-1)*k!/(1*5*9*...*(4*k + 1)). [added 19 Feb 2024: the general result can be proved by the WZ method as described in Wilf.]
In the limit as r -> oo we find that the constant equals Sum_{k >= 0} 2^(k-1)*k!/(Product_{i = 0..k} 4*i + 1) as noted above. (End)
From Peter Bala, Mar 03 2024: (Start)
Continued fraction: 1/(1 + 1^2/(4 + 5^2/(4 + 9^2/(4 + 13^2/(4 + ... ))))) due to Euler.
Equals hypergeom([1/4, 1], [5/4], -1).
Gauss's continued fraction: 1/(1 + 1^2/(5 + 4^2/(9 + 5^2/(13 + 8^2/(17 + 9^2/(21 + 12^2/(25 + 13^2/(29 + 16^2/(33 + 17^2/(37 + ... )))))))))). (End)

A103712 Decimal expansion of the expected distance from a randomly selected point in the unit square to its center: (sqrt(2) + log(1 + sqrt(2)))/6.

Original entry on oeis.org

3, 8, 2, 5, 9, 7, 8, 5, 8, 2, 3, 2, 1, 0, 6, 3, 4, 5, 6, 7, 2, 3, 8, 3, 0, 0, 8, 1, 9, 8, 2, 4, 8, 3, 9, 7, 9, 3, 2, 9, 7, 2, 0, 3, 3, 9, 3, 9, 7, 6, 3, 9, 1, 3, 9, 8, 8, 3, 2, 9, 2, 2, 4, 4, 4, 0, 6, 8, 4, 9, 4, 3, 7, 8, 0, 6, 8, 8, 8, 5, 4, 4, 4, 7, 3, 4, 9, 0, 7, 1, 0, 3, 9, 6, 4, 9, 6, 0, 2, 5, 9, 8, 6, 2, 5
Offset: 0

Views

Author

Sylvester Reese and Jonathan Sondow, Feb 13 2005

Keywords

Comments

Is it a coincidence that this constant is equal to 1/6 of the universal parabolic constant A103710? (Reese, 2004; Finch, 2012)
exp(d(2)) - exp(d(2))/Pi = 0.9994179247351742... ~ 1 - 1/1718. - Gerald McGarvey, Feb 21 2005
Take a point on a line of irrational slope and a line segment of a given length centered at the point, integrate the distance of a point on the line to the set of lattice points along the line segment, and divide by the length. The limit as the length approaches infinity can be shown by a generalization of the Equidistribution Theorem to give the expected distance of a point in the unit square to its corners, this constant. - Thomas Anton, Jun 19 2021

Examples

			0.38259785823210634567238300819824839793297203393976391398832922444...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, section 8.1.
  • S. Reese, A universal parabolic constant, 2004, preprint.

Crossrefs

Equal to (A002193 + A091648)/6 = (A103710)/6 = (A103711)/3.
Cf. A244921.

Programs

  • Mathematica
    RealDigits[(Sqrt[2] + Log[1 + Sqrt[2]])/6, 10, 111][[1]] (* Robert G. Wilson v, Feb 14 2005 *)
  • Maxima
    fpprec: 100$ ev(bfloat((sqrt(2) + log(1 + sqrt(2)))/6)); /* Martin Ettl, Oct 17 2012 */
    
  • PARI
    (sqrt(2) + log(1 + sqrt(2)))/6 \\ G. C. Greubel, Sep 22 2017

Formula

Equals (1/3)*Integral_{x = 0..1} sqrt(1 + x^2) dx. - Peter Bala, Feb 28 2019
Equals Integral_{x>=1} arcsinh(x)/x^4 dx. - Amiram Eldar, Jun 26 2021
Equals A244921 / 2. - Amiram Eldar, Jun 04 2023

A196525 Decimal expansion of log(1+sqrt(2))/sqrt(2).

Original entry on oeis.org

6, 2, 3, 2, 2, 5, 2, 4, 0, 1, 4, 0, 2, 3, 0, 5, 1, 3, 3, 9, 4, 0, 2, 0, 0, 8, 0, 2, 5, 0, 5, 6, 8, 0, 0, 2, 6, 5, 0, 6, 9, 5, 3, 1, 2, 3, 4, 6, 5, 6, 7, 2, 5, 2, 8, 9, 8, 7, 1, 4, 7, 7, 6, 0, 9, 6, 1, 7, 0, 0, 0, 4, 5, 4, 7, 0, 1, 4, 1, 8, 0, 4, 6, 7, 6, 6, 9, 0, 7, 3, 2, 3, 5, 6, 2, 6, 6
Offset: 0

Views

Author

R. J. Mathar, Oct 03 2011

Keywords

Examples

			0.6232252401402305133940200802505680... = A091648/A002193.
From _Peter Bala_, Dec 01 2021: (Start)
With N = 10000, the truncated series Sum_{k = 0..N/4 - 1} (-1)^k/((4*k + 1)*(4*k+3)) = 0.6232252[3]014023[16]1339[3659]080... to 27 decimal places. The square bracketed numbers show where this decimal expansion differs from that of (1/sqrt(2))*log(1+sqrt(2)) = 0.6232252(4)014023(05) 1339(4020)080.... The numbers 1, -11, 361 must be added to the square bracketed numbers to give the correct decimal expansion to 27 decimal places. (End)
		

Crossrefs

Programs

  • Magma
    SetDefaultRealField(RealField(100)); Log(Sqrt(2)+1)/Sqrt(2); // G. C. Greubel, Oct 05 2018
  • Mathematica
    RealDigits[Log[1+Sqrt[2]]/Sqrt[2],10,120][[1]] (* Harvey P. Dale, Dec 27 2011 *)
    RealDigits[Sum[1/((2 n - 1) 2^n), {n, 1, Infinity}], 10, 120][[1]] (* Fred Daniel Kline, May 23 2019 *)
  • PARI
    log(sqrt(2)+1)/sqrt(2) \\ Michel Marcus, Sep 27 2017
    

Formula

Equals Sum_{n>=1} A091337(n)/n = 1 - 1/3 - 1/5 + 1/7 + 1/9 - 1/11 - ...
Equals 2*Sum_{n>=1} (-1)^n/A001539(n). - Michel Marcus, Sep 27 2017
From Fred Daniel Kline, May 23 2019: (Start)
Equals arcsinh(1)/sqrt(2).
Equals Sum_{n>=1} 1/A118417(n-1) = Sum_{n>=1} 1/((2*n - 1)*2^n). (End)
From Peter Bala, Nov 01 2019: (Start)
Equals (1/sqrt(2))*arccoth(sqrt(2)).
Equals 1 - 8*Sum_{n >= 0} (-1)^(n+1)*n/(16*n^2 - 1).
Equals 1 - Integral_{x = 0..inf} exp(-2*x)*cosh(x)/cosh(2*x) dx.
Equals 2*Integral_{x = 0..inf} exp(x)*(exp(2*x) + 1)*(exp(4*x) - 1)/(exp(4*x) + 1)^2 dx - 1. (End)
From Amiram Eldar, Aug 16 2020: (Start)
Equals Sum_{k>=0} (-1)^k * (2*k)!!/(2*k+1)!!.
Equals Integral_{x=0..Pi/4} 1/(cos(x) + sin(x)) dx. (End)
From Peter Bala, Dec 01 2021: (Start)
Equals 2*Sum_{k >= 0} (-1)^k/((4*k + 1)*(4*k + 3)).
Let N be a positive integer divisible by 4. We have the asymptotic expansion (1/sqrt(2))*log(1 + sqrt(2)) - 2*Sum_{k = 0..N/4 - 1} (-1)^k/((4*k + 1)*(4*k + 3)) ~ 1/N^2 - 11/N^4 + 361/N^6 - 24611/N^8 + ..., where the sequence of unsigned coefficients [1, 11, 361, 24611, ...] is A000464. See A181048 and A181049. An example is given below. (End)
Equals 1/Product_{p prime} (1 - Kronecker(8,p)/p), where Kronecker(8,p) = 0 if p = 2, 1 if p == 1 or 7 (mod 8) or -1 if p == 3 or 5 (mod 8). - Amiram Eldar, Dec 17 2023
Equals integral_{x=0..Pi/2} sin^2(x)/(sin(x)+cos(x)) dx [Nahin]. - R. J. Mathar, May 16 2024

A103711 Decimal expansion of the ratio of the length of the latus rectum arc of any parabola to its latus rectum: (sqrt(2) + log(1 + sqrt(2)))/2.

Original entry on oeis.org

1, 1, 4, 7, 7, 9, 3, 5, 7, 4, 6, 9, 6, 3, 1, 9, 0, 3, 7, 0, 1, 7, 1, 4, 9, 0, 2, 4, 5, 9, 4, 7, 4, 5, 1, 9, 3, 7, 9, 8, 9, 1, 6, 1, 0, 1, 8, 1, 9, 2, 9, 1, 7, 4, 1, 9, 6, 4, 9, 8, 7, 6, 7, 3, 3, 2, 2, 0, 5, 4, 8, 3, 1, 3, 4, 2, 0, 6, 6, 5, 6, 3, 3, 4, 2, 0, 4, 7, 2, 1, 3, 1, 1, 8, 9, 4, 8, 8, 0, 7, 7, 9, 5, 8, 7
Offset: 1

Views

Author

Sylvester Reese and Jonathan Sondow, Feb 13 2005

Keywords

Comments

Just as all circles are similar, all parabolas are similar. Just as the ratio of a semicircle to its diameter is always Pi/2, the ratio of the length of the latus rectum arc of any parabola to its latus rectum is (sqrt(2) + log(1 + sqrt(2)))/2.
Let c = this constant and a = e - exp((c+Pi)/2 - log(Pi)), then a = .0000999540234051652627... and c - 10*(-log(exp(a) - a - 1) - 19) = .000650078964115564700067717... - Gerald McGarvey, Feb 21 2005
Half the universal parabolic constant A103710 (the ratio of the length of the latus rectum arc of any parabola to its focal parameter). Like Pi, it is transcendental.
Is it a coincidence that this constant is equal to 3 times the expected distance A103712 from a randomly selected point in the unit square to its center? (Reese, 2004; Finch, 2012)

Examples

			1.14779357469631903701714902459474519379891610181929174196498767332...
		

References

  • H. Dörrie, 100 Great Problems of Elementary Mathematics, Dover, 1965, Problems 57 and 58.
  • C. E. Love, Differential and Integral Calculus, 4th ed., Macmillan, 1950, pp. 286-288.
  • C. S. Ogilvy, Excursions in Geometry, Oxford Univ. Press, 1969, p. 84.
  • S. Reese, A universal parabolic constant, 2004, preprint.

Crossrefs

Equal to (A103710)/2 = (A002193 + A091648)/2 = 3*(A103712).

Programs

  • Mathematica
    RealDigits[(Sqrt[2] + Log[1 + Sqrt[2]])/2, 10, 111][[1]] (* Robert G. Wilson v, Feb 14 2005 *)
    N[Integrate[Sqrt[1 + x^2], {x, 0, 1}], 120] (* Clark Kimberling, Jan 06 2014 *)

Formula

Equals Integral_{x = 0..1} sqrt(1 + x^2) dx. - Peter Bala, Feb 28 2019
Equals Sum_{n>=0} (-1)^(n + 1)*binomial(2*n, n)/((4*n^2 - 1)*4^n). - Antonio Graciá Llorente, Dec 16 2024

A222362 Decimal expansion of the ratio of the area of the latus rectum segment of any equilateral hyperbola to the square of its semi-axis: sqrt(2) - log(1 + sqrt(2)).

Original entry on oeis.org

5, 3, 2, 8, 3, 9, 9, 7, 5, 3, 5, 3, 5, 5, 2, 0, 2, 3, 5, 6, 9, 0, 7, 9, 3, 9, 9, 2, 2, 9, 9, 0, 5, 7, 6, 9, 5, 4, 1, 5, 1, 1, 5, 4, 7, 1, 1, 5, 3, 1, 2, 6, 6, 2, 4, 2, 3, 3, 8, 4, 1, 2, 9, 3, 3, 7, 3, 5, 5, 2, 9, 4, 2, 4, 0, 0, 8, 0, 9, 5, 1, 0, 1, 6, 6, 8, 0, 6, 4, 2, 4, 1, 7, 3, 8, 5, 5, 2, 9, 8, 7, 8, 2, 7, 4, 0, 3, 0, 0, 3
Offset: 0

Views

Author

Sylvester Reese and Jonathan Sondow, Mar 01 2013

Keywords

Comments

Just as circles are ellipses whose semi-axes are equal (and are called the radius of the circle), equilateral (or rectangular) hyperbolas are hyperbolas whose semi-axes are equal.
Just as the ratio of the area of a circle to the square of its radius is always Pi, the ratio of the area of the latus rectum segment of any equilateral hyperbola to the square of its semi-axis is the universal equilateral hyperbolic constant sqrt(2) - log(1 + sqrt(2)).
Note the remarkable similarity to sqrt(2) + log(1 + sqrt(2)), the universal parabolic constant A103710, which is a ratio of arc lengths rather than of areas. Lockhart (2012) says "the arc length integral for the parabola ... is intimately connected to the hyperbolic area integral ... I think it is surprising and wonderful that the length of one conic section is related to the area of another".
This constant is also the abscissa of the vertical asymptote of the involute of the logarithmic curve (starting point (1,0)). - Jean-François Alcover, Nov 25 2016

Examples

			0.532839975353552023569079399229905769541511547115312662423384129337355...
		

References

  • H. Dörrie, 100 Great Problems of Elementary Mathematics, Dover, 1965, Problems 57 and 58.
  • P. Lockhart, Measurement, Harvard University Press, 2012, p. 369.

Crossrefs

Programs

Formula

Sqrt(2) - arcsinh(1), also equals Integral_{1..oo} 1/(x^2*(1+x)^(1/2)) dx. - Jean-François Alcover, Apr 16 2015
Equals Integral_{x = 0..1} x^2/sqrt(1 + x^2) dx. - Peter Bala, Feb 28 2019

A257435 Decimal expansion of G(1/6), a generalized Catalan constant.

Original entry on oeis.org

9, 0, 0, 4, 2, 4, 6, 0, 0, 3, 8, 9, 7, 0, 7, 7, 5, 7, 8, 5, 8, 8, 2, 7, 5, 8, 9, 0, 2, 9, 0, 4, 9, 4, 8, 5, 8, 2, 9, 9, 4, 3, 9, 5, 7, 6, 6, 6, 6, 1, 8, 7, 6, 5, 5, 9, 5, 1, 5, 7, 3, 1, 8, 3, 9, 1, 0, 5, 4, 4, 2, 0, 3, 6, 7, 5, 6, 5, 4, 7, 4, 9, 9, 6, 2, 3, 2, 3, 1, 5, 3, 0, 2, 5, 7, 1, 2, 4, 8, 2, 2, 8, 7, 8, 6
Offset: 0

Views

Author

Jean-François Alcover, Apr 23 2015

Keywords

Examples

			0.900424600389707757858827589029049485829943957666618765595157318391...
		

Crossrefs

Cf. A006752 (G(0) = Catalan), A091648 (G(1/4)), A257436 (G(1/3)), A257437 (G(1/12)), A257438 (G(1/5)).

Programs

  • Magma
    SetDefaultRealField(RealField(100)); (3/4)*Sqrt(3)*Log(2); // G. C. Greubel, Aug 24 2018
  • Mathematica
    RealDigits[(3/4)*Sqrt[3]*Log[2], 10, 105] // First
    N[Pi*HypergeometricPFQ[{1/3, 1/2, 2/3}, {1, 3/2}, 1]/4, 105] (* Vaclav Kotesovec, Apr 24 2015 *)
  • PARI
    (3/4)*sqrt(3)*log(2) \\ G. C. Greubel, Aug 24 2018
    

Formula

G(s) = (Pi/4) * 3F2(1/2, 1/2-s, s+1/2; 1, 3/2; 1), with 2F1 the hypergeometric function.
G(s) = (1/(8*s))*(Pi + cos(Pi*s)*(psi(1/4 + s/2) - psi(3/4 + s/2))), where psi is the digamma function (PolyGamma).
G(1/6) = (3/4)*sqrt(3)*log(2).
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