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 1-4 of 4 results.

A085565 Decimal expansion of lemniscate constant A.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Jul 06 2003

Keywords

Comments

This number is transcendental by a result of Schneider on elliptic integrals. - Benoit Cloitre, Jan 08 2006
The two lemniscate constants are A = Integral_{x = 0..1} 1/sqrt(1 - x^4) dx and B = Integral_{x = 0..1} x^2/sqrt(1 - x^4) dx. See A076390. - Peter Bala, Oct 25 2019
Also the ratio of generating curve length to diameter of a "Mylar balloon" (see Paulsen). - Jeremy Tan, May 05 2021

Examples

			1.3110287771460599052324197949455597068413774757158115814084108519...
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.
  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 6.1, p. 421.
  • Th. Schneider, Transzendenzuntersuchungen periodischer Funktionen (1934).
  • Th. Schneider, Arithmetische Untersuchungen elliptischer Integrale (1937).

Crossrefs

Programs

  • Magma
    C := ComplexField(); [Gamma(1/4)^2/(4*Sqrt(2*Pi(C)))]; // G. C. Greubel, Nov 05 2017
  • Mathematica
    RealDigits[ Gamma[1/4]^2/(4*Sqrt[2*Pi]), 10, 99][[1]]
    (* or *)
    RealDigits[ EllipticK[-1], 10, 99][[1]] (* Jean-François Alcover, Mar 07 2013, updated Jul 30 2016 *)
  • PARI
    gamma(1/4)^2/4/sqrt(2*Pi)
    
  • PARI
    K(x)=Pi/2/agm(1,sqrt(1-x))
    K(-1) \\ Charles R Greathouse IV, Aug 02 2018
    
  • PARI
    ellK(I) \\ Charles R Greathouse IV, Feb 04 2025
    

Formula

Equals (1/4)*(2*Pi)^(-1/2)*GAMMA(1/4)^2.
Equals Integral_{x>=1}dx/sqrt(4x^3-4x). - Benoit Cloitre, Jan 08 2006
Equals Product_(k>=0, [(4k+3)(4k+4)] / [(4k+5)(4k+2)] ) (Gauss). - Ralf Stephan, Mar 04 2008 [corrected by Vaclav Kotesovec, May 01 2020]
Equals Pi/sqrt(8)/agm(1,sqrt(1/2)).
Equals Pi/sqrt(8)*hypergeom([1/2,1/2],[1],1/2).
Product_{m>=1} ((2*m)/(2*m+1))^(-1)^m. - Jean-François Alcover, Sep 02 2014, after Steven Finch
From Peter Bala, Mar 09 2015: (Start)
Equals Integral_{x = 0..1} 1/sqrt(1 - x^4) dx.
Continued fraction representations: 2/(1 + 1*3/(2 + 5*7/(2 + 9*11/(2 + ... )))) due to Euler - see Khrushchev, p. 179.
Also equals 1 + 1/(2 + 2*3/(2 + 4*5/(2 + 6*7/(2 + ... )))). (End)
From Peter Bala, Oct 25 2019: (Start)
Equals 1 + 1/5 + (1*3)/(5*9) + (1*3*5)/(5*9*13) + ... = hypergeom([1/2,1],[5/4],1/2) by Gauss's second summation theorem.
Equivalently, define a sequence of rational numbers r(n) recursively by r(n) = (2*n - 3)/(4*n - 3)*r(n-1) with r(1) = 1. Then the constant equals Sum_{n >= 1} r(n) = 1 + 1/5 + 1/15 + 1/39 + 7/663 + 1/221 + 11/5525 + 11/12325 + 1/2465 + .... The partial sum of the series to 100 terms gives 32 correct decimal digits for the constant.
Equals (1*3)/(1*5) + (1*3*5)/(1*5*9) + (1*3*5*7)/(1*5*9*13) + ... = (3/5) * hypergeom([5/2,1],[9/4],1/2). (End)
Equals (3/2)*A225119. - Peter Bala, Oct 27 2019
Equals Integral_{x=0..Pi/2} 1/sqrt(1 + cos(x)^2) dx = Integral_{x=0..Pi/2} 1/sqrt(1 + sin(x)^2) dx. - Amiram Eldar, Aug 09 2020
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 = 4*k + 1 for k >= 0.
For example, taking k = 0 and k = 1 yields
A = 2/(1 + (1*3)/(2 + (5*7)/(2 + (9*11)/(2 + (13*15)/(2 + ... + (4*n + 1)*(4*n + 3)/(2 + ... )))))) and
A = (1/4)*(5 + (1*3)/(10 + (5*7)/(10 + (9*11)/(10 + (13*15)/(10 + ... + (4*n + 1)*(4*n + 3)/(10 + ... )))))). (End)

A243340 Decimal expansion of 4*L/(3*Pi), a constant related to the asymptotic evaluation of the number of primes of the form a^2+b^4, where L is Gauss' lemniscate constant.

Original entry on oeis.org

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

Author

Jean-François Alcover, Jun 03 2014

Keywords

Examples

			1.11283578889876424837523964373206241199199...
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.
  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 2.3 Landau-Ramanujan constant, p. 102.

Crossrefs

Cf. A062539 (L), A076390, A085565, A225119 (L/3).

Programs

  • Mathematica
    L = Pi^(3/2)/(Sqrt[2]*Gamma[3/4]^2); RealDigits[4*L/(3*Pi), 10, 103] // First
  • PARI
    2*sqrt(2*Pi)/(3*gamma(3/4)^2) \\ Stefano Spezia, Nov 27 2024

Formula

Equals 2*sqrt(2*Pi)/(3*Gamma(3/4)^2).
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 = 4*k + 3 for k >= 0.
For example, taking k = 0 and k = 1 yields
4*L/(3*Pi) = 1 + 1/(6 + (5*7)/(6 + (9*11)/(6 + (13*15)/(6 + ... + (4*n + 1)*(4*n + 3)/(6 + ... ))))) and
4*L/(3*Pi) = 8/(7 + (1*3)/(14 + (5*7)/(14 + (9*11)/(14 + (13*15)/(14 + ... + (4*n + 1)*(4*n + 3)/(14 + ... )))))).
Equals (2/3) * 1/A076390. (End)

A281792 Primes of the form x^2 + p^4 where x > 0 and p is prime.

Original entry on oeis.org

17, 41, 97, 137, 181, 241, 277, 337, 457, 641, 661, 757, 769, 821, 857, 881, 977, 1109, 1201, 1237, 1301, 1409, 1697, 2017, 2069, 2389, 2417, 2437, 2617, 2657, 2741, 2801, 3041, 3217, 3301, 3329, 3541, 3557, 3697, 3761, 3989, 4001, 4177, 4241, 4337, 4517, 4721, 5557, 5641, 5857, 6101, 6257, 6481, 6577
Offset: 1

Author

Robert Israel, Jan 30 2017

Keywords

Comments

Heath-Brown and Li prove an asymptotic formula for the number of terms <= x, in particular showing that the sequence is infinite.

Examples

			17 = 1^2 + 2^4
41 = 5^2 + 2^4
97 = 9^2 + 2^4
137 = 11^2 + 2^4
181 = 10^2 + 3^4
		

Crossrefs

Subsequence of A028916.

Programs

  • Maple
    N:= 10000: # to get all terms <= N
    A:= select(isprime, {seq(seq(x^2+y^4, x=1..floor(sqrt(N-y^4))),
    y=select(isprime, [$1..floor(N^(1/4))]))}):
    sort(convert(A,list)); # Robert Israel, Jan 30 2017
  • Mathematica
    nn = 10000;
    Select[Table[x^2+y^4, {y, Select[Range[nn^(1/4)], PrimeQ]}, {x, Sqrt[nn-y^4 ]}] // Flatten, PrimeQ] // Union (* Jean-François Alcover, Sep 18 2018, after Robert Israel *)
  • PARI
    list(lim)=if(lim<17, return([])); my(v=List(),p4,t); forstep(a=1,sqrtint(-16+lim\=1),2, if(isprime(t=a^2+16), listput(v,t))); forprime(p=3,sqrtnint(lim-4,4), p4=p^4; forstep(a=2,sqrtint(lim-p4),2, if(isprime(t=p4+a^2), listput(v,t)))); Set(v) \\ Charles R Greathouse IV, Aug 13 2017

Formula

Heath-Brown and Li prove that there are c*x^(3/4)/log^2 x terms up to x, where c = 4*nu*J = 4.79946121442200811438003177..., nu = A199401, and J = A225119. - Charles R Greathouse IV, Aug 21 2017

A371861 Decimal expansion of Integral_{x=0..1} sqrt(1 - x^3) dx.

Original entry on oeis.org

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

Author

Ilya Gutkovskiy, Apr 09 2024

Keywords

Examples

			0.8413092631952725567050114474301764812778...
		

Crossrefs

Decimal expansions of Integral_{x=0..1} sqrt(1 - x^k) dx: A003881 (k=2), this sequence (k=3), A225119 (k=4).

Programs

  • Mathematica
    RealDigits[Sqrt[Pi] Gamma[1/3]/(6 Gamma[11/6]), 10, 103][[1]]
    RealDigits[Sqrt[3] * Gamma[1/3]^3 / (5*Pi*2^(4/3)), 10, 103][[1]] (* Vaclav Kotesovec, Apr 09 2024 *)
  • PARI
    intnum(x=0, 1, sqrt(1 - x^3)) \\ Michel Marcus, Apr 10 2024

Formula

Equals sqrt(Pi) * Gamma(1/3) / (6 * Gamma(11/6)).
Equals sqrt(3) * Gamma(1/3)^3 / (5*Pi*2^(4/3)). - Vaclav Kotesovec, Apr 09 2024
Equals 3*A118292/10. - Hugo Pfoertner, Apr 09 2024
Showing 1-4 of 4 results.