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 16 results. Next

A131681 Phi(binomial(2*n,n)*n^2) (phi of A002736 = Apery numbers: n^2 C(2n,n). ).

Original entry on oeis.org

0, 1, 8, 48, 384, 1440, 8640, 40320, 184320, 829440, 5529600, 18247680, 87588864, 474439680, 1532805120, 7664025600, 32699842560, 130288435200, 517321728000, 2184247296000, 9196830720000, 38626689024000, 188841590784000
Offset: 0

Views

Author

Zerinvary Lajos, Oct 07 2007

Keywords

Crossrefs

Cf. A002736.

Programs

  • Maple
    with(numtheory):with(combinat):a:=n->phi(binomial(2*n,n)*n^2): seq(a(n), n=0..25);

A000796 Decimal expansion of Pi (or digits of Pi).

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane

Keywords

Comments

Sometimes called Archimedes's constant.
Ratio of a circle's circumference to its diameter.
Also area of a circle with radius 1.
Also surface area of a sphere with diameter 1.
A useful mnemonic for remembering the first few terms: How I want a drink, alcoholic of course, after the heavy lectures involving quantum mechanics ...
Also ratio of surface area of sphere to one of the faces of the circumscribed cube. Also ratio of volume of a sphere to one of the six inscribed pyramids in the circumscribed cube. - Omar E. Pol, Aug 09 2012
Also surface area of a quarter of a sphere of radius 1. - Omar E. Pol, Oct 03 2013
Also the area under the peak-shaped even function f(x)=1/cosh(x). Proof: for the upper half of the integral, write f(x) = (2*exp(-x))/(1+exp(-2x)) = 2*Sum_{k>=0} (-1)^k*exp(-(2k+1)*x) and integrate term by term from zero to infinity. The result is twice the Gregory series for Pi/4. - Stanislav Sykora, Oct 31 2013
A curiosity: a 144 X 144 magic square of 7th powers was recently constructed by Toshihiro Shirakawa. The magic sum = 3141592653589793238462643383279502884197169399375105, which is the concatenation of the first 52 digits of Pi. See the MultiMagic Squares link for details. - Christian Boyer, Dec 13 2013 [Comment revised by N. J. A. Sloane, Aug 27 2014]
x*Pi is also the surface area of a sphere whose diameter equals the square root of x. - Omar E. Pol, Dec 25 2013
Also diameter of a sphere whose surface area equals the volume of the circumscribed cube. - Omar E. Pol, Jan 13 2014
From Daniel Forgues, Mar 20 2015: (Start)
An interesting anecdote about the base-10 representation of Pi, with 3 (integer part) as first (index 1) digit:
358 0
359 3
360 6
361 0
362 0
And the circle is customarily subdivided into 360 degrees (although Pi radians yields half the circle)...
(End)
Sometimes referred to as Archimedes's constant, because the Greek mathematician computed lower and upper bounds of Pi by drawing regular polygons inside and outside a circle. In Germany it was called the Ludolphian number until the early 20th century after the Dutch mathematician Ludolph van Ceulen (1540-1610), who calculated up to 35 digits of Pi in the late 16th century. - Martin Renner, Sep 07 2016
As of the beginning of 2019 more than 22 trillion decimal digits of Pi are known. See the Wikipedia article "Chronology of computation of Pi". - Harvey P. Dale, Jan 23 2019
On March 14, 2019, Emma Haruka Iwao announced the calculation of 31.4 trillion digits of Pi using Google Cloud's infrastructure. - David Radcliffe, Apr 10 2019
Also volume of three quarters of a sphere of radius 1. - Omar E. Pol, Aug 16 2019
On August 5, 2021, researchers from the University of Applied Sciences of the Grisons in Switzerland announced they had calculated 62.8 trillion digits. Guinness World Records has not verified this yet. - Alonso del Arte, Aug 23 2021
The Hermite-Lindemann (1882) theorem states, that if z is a nonzero algebraic number, then e^z is a transcendent number. The transcendence of Pi then results from Euler's relation: e^(i*Pi) = -1. - Peter Luschny, Jul 21 2023
The 10000 words of the book "Not A Wake" by Michael Keith, written in Pilish, match in length the first 10000 digits of Pi. - Paolo Xausa, Aug 07 2025

Examples

			3.1415926535897932384626433832795028841971693993751058209749445923078164062\
862089986280348253421170679821480865132823066470938446095505822317253594081\
284811174502841027019385211055596446229489549303819...

References

  • Mohammad K. Azarian, A Summary of Mathematical Works of Ghiyath ud-din Jamshid Kashani, Journal of Recreational Mathematics, Vol. 29(1), pp. 32-42, 1998.
  • J. Arndt & C. Haenel, Pi Unleashed, Springer NY 2001.
  • P. Beckmann, A History of Pi, Golem Press, Boulder, CO, 1977.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 396.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 237-239.
  • J.-P. Delahaye, Le fascinant nombre pi, Pour la Science, Paris 1997.
  • P. Eyard and J.-P. Lafon, The Number Pi, Amer. Math. Soc., 2004.
  • S. R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, Section 1.4.
  • Le Petit Archimede, Special Issue On Pi, Supplement to No. 64-5, May 1980 ADCS Amiens.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 31.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 1, equations 1:7:1, 1:7:2 at pages 12-13.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See pp. 48-55.

Links

Crossrefs

Cf. A001203 (continued fraction).
Pi in base b: A004601 (b=2), A004602 (b=3), A004603 (b=4), A004604 (b=5), A004605 (b=6), A004606 (b=7), A006941 (b=8), A004608 (b=9), this sequence (b=10), A068436 (b=11), A068437 (b=12), A068438 (b=13), A068439 (b=14), A068440 (b=15), A062964 (b=16), A224750 (b=26), A224751 (b=27), A060707 (b=60). - Jason Kimberley, Dec 06 2012
Decimal expansions of expressions involving Pi: A002388 (Pi^2), A003881 (Pi/4), A013661 (Pi^2/6), A019692 (2*Pi=tau), A019727 (sqrt(2*Pi)), A059956 (6/Pi^2), A060294 (2/Pi), A091925 (Pi^3), A092425 (Pi^4), A092731 (Pi^5), A092732 (Pi^6), A092735 (Pi^7), A092736 (Pi^8), A163973 (Pi/log(2)).
Cf. A001901 (Pi/2; Wallis), A002736 (Pi^2/18; Euler), A007514 (Pi), A048581 (Pi; BBP), A054387 (Pi; Newton), A092798 (Pi/2), A096954 (Pi/4; Machin), A097486 (Pi), A122214 (Pi/2), A133766 (Pi/4 - 1/2), A133767 (5/6 - Pi/4), A166107 (Pi; MGL).
Cf. A001622, A048940, A248682.

Programs

  • Haskell
    -- see link: Literate Programs
import Data.Char (digitToInt)
a000796 n = a000796_list (n + 1) !! (n + 1)
a000796_list len = map digitToInt $ show $ machin' `div` (10 ^ 10) where
   machin' = 4 * (4 * arccot 5 unity - arccot 239 unity)
   unity = 10 ^ (len + 10)
   arccot x unity = arccot' x unity 0 (unity `div` x) 1 1 where
     arccot' x unity summa xpow n sign
      | term == 0 = summa
      | otherwise = arccot'
        x unity (summa + sign * term) (xpow `div` x ^ 2) (n + 2) (- sign)
      where term = xpow `div` n
-- Reinhard Zumkeller, Nov 24 2012
  • Haskell
    -- See Niemeijer link and also Gibbons link.
a000796 n = a000796_list !! (n-1) :: Int
a000796_list = map fromInteger $ piStream (1, 0, 1)
   [(n, a*d, d) | (n, d, a) <- map (\k -> (k, 2 * k + 1, 2)) [1..]] where
   piStream z xs'@(x:xs)
     | lb /= approx z 4 = piStream (mult z x) xs
     | otherwise = lb : piStream (mult (10, -10 * lb, 1) z) xs'
     where lb = approx z 3
           approx (a, b, c) n = div (a * n + b) c
           mult (a, b, c) (d, e, f) = (a * d, a * e + b * f, c * f)
-- Reinhard Zumkeller, Jul 14 2013, Jun 12 2013
  • Macsyma
    py(x) := if equal(6,6+x^2) then 2*x else (py(x:x/3),3*%%-4*(%%-x)^3); py(3.); py(dfloat(%)); block([bfprecision:35], py(bfloat(%))) /* Bill Gosper, Sep 09 2002 */
  • Magma
    pi:=Pi(RealField(110)); Reverse(Intseq(Floor(10^105*pi))); // Bruno Berselli, Mar 12 2013
  • Maple
    Digits := 110: Pi*10^104:
ListTools:-Reverse(convert(floor(%), base, 10)); # Peter Luschny, Oct 29 2019
  • Mathematica
    RealDigits[ N[ Pi, 105]] [[1]]
Table[ResourceFunction["NthDigit"][Pi, n], {n, 1, 102}] (* Joan Ludevid, Jun 22 2022; easy to compute a(10000000)=7 with this function; requires Mathematica 12.0+ *)
  • PARI
    { default(realprecision, 20080); x=Pi; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b000796.txt", n, " ", d)); } \\ Harry J. Smith, Apr 15 2009
  • PARI
    A796=[]; A000796(n)={if(n>#A796, localprec(n*6\5+29); A796=digits(Pi\.1^(precision(Pi)-3))); A796[n]} \\ NOTE: as the other programs, this returns the n-th term of the sequence, with n = 1, 2, 3, ... and not n = 1, 0, -1, -2, .... - M. F. Hasler, Jun 21 2022
  • PARI
    first(n)= default(realprecision, n+10); digits(floor(Pi*10^(n-1))) \\ David A. Corneth, Jun 21 2022
  • PARI
    lista(nn, p=20)= {my(u=10^(nn+p+1), f(x, u)=my(n=1, q=u\x, r=q, s=1, t); while(t=(q\=(x*x))\(n+=2), r+=(s=-s)*t); r*4); digits((4*f(5, u)-f(239, u))\10^(p+2)); } \\ Machin-like, with p > the maximal number of consecutive 9-digits to be expected (A048940) - Ruud H.G. van Tol, Dec 26 2024
  • Python
    from sympy import pi, N; print(N(pi, 1000)) # David Radcliffe, Apr 10 2019
  • Python
    from mpmath import mp
def A000796(n):
    if n >= len(A000796.str): mp.dps = n*6//5+50; A000796.str = str(mp.pi-5/mp.mpf(10)**mp.dps)
    return int(A000796.str[n if n>1 else 0])
A000796.str = '' # M. F. Hasler, Jun 21 2022
  • SageMath
    m=125
x=numerical_approx(pi, digits=m+5)
a=[ZZ(i) for i in x.str(skip_zeroes=True) if i.isdigit()]
a[:m] # G. C. Greubel, Jul 18 2023

Formula

Pi = 4*Sum_{k>=0} (-1)^k/(2k+1) [Madhava-Gregory-Leibniz, 1450-1671]. - N. J. A. Sloane, Feb 27 2013
From Johannes W. Meijer, Mar 10 2013: (Start)
2/Pi = (sqrt(2)/2) * (sqrt(2 + sqrt(2))/2) * (sqrt(2 + sqrt(2 + sqrt(2)))/2) * ... [Viete, 1593]
2/Pi = Product_{k>=1} (4*k^2-1)/(4*k^2). [Wallis, 1655]
Pi = 3*sqrt(3)/4 + 24*(1/12 - Sum_{n>=2} (2*n-2)!/((n-1)!^2*(2*n-3)*(2*n+1)*2^(4*n-2))). [Newton, 1666]
Pi/4 = 4*arctan(1/5) - arctan(1/239). [Machin, 1706]
Pi^2/6 = 3*Sum_{n>=1} 1/(n^2*binomial(2*n,n)). [Euler, 1748]
1/Pi = (2*sqrt(2)/9801) * Sum_{n>=0} (4*n)!*(1103+26390*n)/((n!)^4*396^(4*n)). [Ramanujan, 1914]
1/Pi = 12*Sum_{n>=0} (-1)^n*(6*n)!*(13591409 + 545140134*n)/((3*n)!*(n!)^3*(640320^3)^(n+1/2)). [David and Gregory Chudnovsky, 1989]
Pi = Sum_{n>=0} (1/16^n) * (4/(8*n+1) - 2/(8*n+4) - 1/(8*n+5) - 1/(8*n+6)). [Bailey-Borwein-Plouffe, 1989] (End)
Pi = 4 * Sum_{k>=0} 1/(4*k+1) - 1/(4*k+3). - Alexander R. Povolotsky, Dec 25 2008
Pi = 4*sqrt(-1*(Sum_{n>=0} (i^(2*n+1))/(2*n+1))^2). - Alexander R. Povolotsky, Jan 25 2009
Pi = Integral_{x=-oo..oo} dx/(1+x^2). - Mats Granvik and Gary W. Adamson, Sep 23 2012
Pi - 2 = 1/1 + 1/3 - 1/6 - 1/10 + 1/15 + 1/21 - 1/28 - 1/36 + 1/45 + ... [Jonas Castillo Toloza, 2007], that is, Pi - 2 = Sum_{n>=1} (1/((-1)^floor((n-1)/2)*(n^2+n)/2)). - José de Jesús Camacho Medina, Jan 20 2014
Pi = 3 * Product_{t=img(r),r=(1/2+i*t) root of zeta function} (9+4*t^2)/(1+4*t^2) <=> RH is true. - Dimitris Valianatos, May 05 2016
From Ilya Gutkovskiy, Aug 07 2016: (Start)
Pi = Sum_{k>=1} (3^k - 1)*zeta(k+1)/4^k.
Pi = 2*Product_{k>=2} sec(Pi/2^k).
Pi = 2*Integral_{x>=0} sin(x)/x dx. (End)
Pi = 2^{k + 1}*arctan(sqrt(2 - a_{k - 1})/a_k) at k >= 2, where a_k = sqrt(2 + a_{k - 1}) and a_1 = sqrt(2). - Sanjar Abrarov, Feb 07 2017
Pi = Integral_{x = 0..2} sqrt(x/(2 - x)) dx. - Arkadiusz Wesolowski, Nov 20 2017
Pi = lim_{n->oo} 2/n * Sum_{m=1,n} ( sqrt( (n+1)^2 - m^2 ) - sqrt( n^2 - m^2 ) ). - Dimitri Papadopoulos, May 31 2019
From Peter Bala, Oct 29 2019: (Start)
Pi = Sum_{n >= 0} 2^(n+1)/( binomial(2*n,n)*(2*n + 1) ) - Euler.
More generally, Pi = (4^x)*x!/(2*x)! * Sum_{n >= 0} 2^(n+1)*(n+x)!*(n+2*x)!/(2*n+2*x+1)! = 2*4^x*x!^2/(2*x+1)! * hypergeom([2*x+1,1], [x+3/2], 1/2), valid for complex x not in {-1,-3/2,-2,-5/2,...}. Here, x! is shorthand notation for the function Gamma(x+1). This identity may be proved using Gauss's second summation theorem.
Setting x = 3/4 and x = -1/4 (resp. x = 1/4 and x = -3/4) in the above identity leads to series representations for the constant A085565 (resp. A076390). (End)
Pi = Im(log(-i^i)) = log(i^i)*(-2). - Peter Luschny, Oct 29 2019
From Amiram Eldar, Aug 15 2020: (Start)
Equals 2 + Integral_{x=0..1} arccos(x)^2 dx.
Equals Integral_{x=0..oo} log(1 + 1/x^2) dx.
Equals Integral_{x=0..oo} log(1 + x^2)/x^2 dx.
Equals Integral_{x=-oo..oo} exp(x/2)/(exp(x) + 1) dx. (End)
Equals 4*(1/2)!^2 = 4*Gamma(3/2)^2. - Gary W. Adamson, Aug 23 2021
From Peter Bala, Dec 08 2021: (Start)
Pi = 32*Sum_{n >= 1} (-1)^n*n^2/((4*n^2 - 1)*(4*n^2 - 9))= 384*Sum_{n >= 1} (-1)^(n+1)*n^2/((4*n^2 - 1)*(4*n^2 - 9)*(4*n^2 - 25)).
More generally, it appears that for k = 1,2,3,..., Pi = 16*(2*k)!*Sum_{n >= 1} (-1)^(n+k+1)*n^2/((4*n^2 - 1)* ... *(4*n^2 - (2*k+1)^2)).
Pi = 32*Sum_{n >= 1} (-1)^(n+1)*n^2/(4*n^2 - 1)^2 = 768*Sum_{n >= 1} (-1)^(n+1)*n^2/((4*n^2 - 1)^2*(4*n^2 - 9)^2).
More generally, it appears that for k = 0,1,2,..., Pi = 16*Catalan(k)*(2*k)!*(2*k+2)!*Sum_{n >= 1} (-1)^(n+1)*n^2/((4*n^2 - 1)^2* ... *(4*n^2 - (2*k+1)^2)^2).
Pi = (2^8)*Sum_{n >= 1} (-1)^(n+1)*n^2/(4*n^2 - 1)^4 = (2^17)*(3^5)*Sum_{n >= 2} (-1)^n*n^2*(n^2 - 1)/((4*n^2 - 1)^4*(4*n^2 - 9)^4) = (2^27)*(3^5)*(5^5)* Sum_{n >= 3} (-1)^(n+1)*n^2*(n^2 - 1)*(n^2 - 4)/((4*n^2 - 1)^4*(4*n^2 - 9)^4*(4*n^2 - 25)^4). (End)
For odd n, Pi = (2^(n-1)/A001818((n-1)/2))*gamma(n/2)^2. - Alan Michael Gómez Calderón, Mar 11 2022
Pi = 4/phi + Sum_{n >= 0} (1/phi^(12*n)) * ( 8/((12*n+3)*phi^3) + 4/((12*n+5)*phi^5) - 4/((12*n+7)*phi^7) - 8/((12*n+9)*phi^9) - 4/((12*n+11)*phi^11) + 4/((12*n+13)*phi^13) ) where phi = (1+sqrt(5))/2. - Chittaranjan Pardeshi, May 16 2022
Pi = sqrt(3)*(27*S - 36)/24, where S = A248682. - Peter Luschny, Jul 22 2022
Equals Integral_{x=0..1} 1/sqrt(x-x^2) dx. - Michal Paulovic, Sep 24 2023
From Peter Bala, Oct 28 2023: (Start)
Pi = 48*Sum_{n >= 0} (-1)^n/((6*n + 1)*(6*n + 3)*(6*n + 5)).
More generally, it appears that for k >= 0 we have Pi = A(k) + B(k)*Sum_{n >= 0} (-1)^n/((6*n + 1)*(6*n + 3)*...*(6*n + 6*k + 5)), where A(k) is a rational approximation to Pi and B(k) = (3 * 2^(3*k+3) * (3*k + 2)!) / (2^(3*k+1) - (-1)^k). The first few values of A(k) for k >= 0 are [0, 256/85, 65536/20955, 821559296/261636375, 6308233216/2008080987, 908209489444864/289093830828075, ...].
Pi = 16/5 - (288/5)*Sum_{n >= 0} (-1)^n * (6*n + 1)/((6*n + 1)*(6*n + 3)*...*(6*n + 9)).
More generally, it appears that for k >= 0 we have Pi = C(k) + D(k)*Sum_{n >= 0} (-1)^n* (6*n + 1)/((6*n + 1)*(6*n + 3)*...*(6*n + 6*k + 3)), where C(k) and D(k) are rational numbers. The case k = 0 is the Madhava-Gregory-Leibniz series for Pi.
Pi = 168/53 + (288/53)*Sum_{n >= 0} (-1)^n * (42*n^2 + 25*n)/((6*n + 1)*(6*n + 3)*(6*n + 5)*(6*n + 7)).
More generally, it appears that for k >= 1 we have Pi = E(k) + F(k)*Sum_{n >= 0} (-1)^n * (6*(6*k + 1)*n^2 + (24*k + 1)*n)/((6*n + 1)*(6*n + 3)*...*(6*n + 6*k + 1)), where E(k) and F(k) are rational numbers. (End)
From Peter Bala, Nov 10 2023: (Start)
The series representation Pi = 4 * Sum_{k >= 0} 1/(4*k + 1) - 1/(4*k + 3) given above by Alexander R. Povolotsky, Dec 25 2008, is the case n = 0 of the more general result (obtained by the WZ method): for n >= 0, there holds
Pi = Sum_{j = 0.. n-1} 2^(j+1)/((2*j + 1)*binomial(2*j,j)) + 8*(n+1)!*Sum_{k >= 0} 1/((4*k + 1)*(4*k + 3)*...*(4*k + 2*n + 3)).
Letting n -> oo gives the rapidly converging series Pi = Sum_{j >= 0} 2^(j+1)/((2*j + 1)*binomial(2*j,j)) due to Euler.
More generally, it appears that for n >= 1, Pi = 1/(2*n-1)!!^2 * Sum_{j >= 0} (Product_{i = 0..2*n-1} j - i) * 2^(j+1)/((2*j + 1)*binomial(2*j,j)).
For any integer n, Pi = (-1)^n * 4 * Sum_{k >= 0} 1/(4*k + 1 + 2*n) - 1/(4*k + 3 - 2*n). (End)
Pi = Product_{k>=1} ((k^3*(k + 2)*(2*k + 1)^2)/((k + 1)^4*(2*k - 1)^2))^k. - Antonio Graciá Llorente, Jun 13 2024
Equals Integral_{x=0..2} sqrt(8 - x^2) dx - 2 (see Ambrisi and Rizzi). - Stefano Spezia, Jul 21 2024
Equals 3 + 4*Sum_{k>0} (-1)^(k+1)/(4*k*(1+k)*(1+2*k)) (see Wells at p. 53). - Stefano Spezia, Aug 31 2024
Equals 4*Integral_{x=0..1} sqrt(1 - x^2) dx = lim_{n->oo} (4/n^2)*Sum_{k=0..n} sqrt(n^2 - k^2) (see Finch). - Stefano Spezia, Oct 19 2024
Equals Beta(1/2,1/2) (see Shamos). - Stefano Spezia, Jun 03 2025
From Kritsada Moomuang, Jun 18 2025: (Start)
Equals 2 + Integral_{x=0..1} 1/(sqrt(x)*(1 + sqrt(1 - x))) dx.
Equals 2 + Integral_{x=0..1} log(1 + sqrt(1 - x))/sqrt(x) dx. (End)
Pi = 2*arccos(1/phi) + arccos(1/phi^3) = 4*arcsin(1/phi) + 2*arcsin(1/phi^3) where phi = (1+sqrt(5))/2. - Chittaranjan Pardeshi, Jul 02 2025
Pi = Sum_{n >= 0} zeta(2*n)*(2^(2*n - 1) - 1)/2^(4*n - 3). - Andrea Pinos, Jul 29 2025

Extensions

Additional comments from William Rex Marshall, Apr 20 2001

A005259 Apery (Apéry) numbers: Sum_{k=0..n} (binomial(n,k)*binomial(n+k,k))^2.

Original entry on oeis.org

1, 5, 73, 1445, 33001, 819005, 21460825, 584307365, 16367912425, 468690849005, 13657436403073, 403676083788125, 12073365010564729, 364713572395983725, 11111571997143198073, 341034504521827105445, 10534522198396293262825, 327259338516161442321485
Offset: 0

Views

Author

Simon Plouffe, N. J. A. Sloane, May 20 1991

Keywords

Comments

Conjecture: For each n = 1,2,3,... the Apéry polynomial A_n(x) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
The expansions of exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + 232771*x^5 + ... and exp( Sum_{n >= 1} a(n-1)*x^n/n ) = 1 + 3*x + 27*x^2 + 390*x^3 + 7038*x^4 + 144550*x^5 + ... both appear to have integer coefficients. See A267220. - Peter Bala, Jan 12 2016
Diagonal of the rational function R(x, y, z, w) = 1 / (1 - (w*x*y*z + w*x*y + w*z + x*y + x*z + y + z)); also diagonal of rational function H(x, y, z, w) = 1/(1 - w*(1+x)*(1+y)*(1+z)*(x*y*z + y*z + y + z + 1)). - Gheorghe Coserea, Jun 26 2018
Named after the French mathematician Roger Apéry (1916-1994). - Amiram Eldar, Jun 10 2021

Examples

			G.f. = 1 + 5*x + 73*x^2 + 1445*x^3 + 33001*x^4 + 819005*x^5 + 21460825*x^6 + ...
a(2) = (binomial(2,0) * binomial(2+0,0))^2 + (binomial(2,1) * binomial(2+1,1))^2 + (binomial(2,2) * binomial(2+2,2))^2 = (1*1)^2 + (2*3)^2 + (1*6)^2 = 1 + 36 + 36 = 73. - _Michael B. Porter_, Jul 14 2016

References

  • Julian Havil, The Irrationals, Princeton University Press, Princeton and Oxford, 2012, pp. 137-153.
  • Wolfram Koepf, Hypergeometric Identities. Ch. 2 in Hypergeometric Summation: An Algorithmic Approach to Summation and Special Function Identities. Braunschweig, Germany: Vieweg, pp. 55, 119 and 146, 1998.
  • Maxim Kontsevich and Don Zagier, Periods, pp. 771-808 of B. Engquist and W. Schmid, editors, Mathematics Unlimited - 2001 and Beyond, 2 vols., Springer-Verlag, 2001.
  • Leonard Lipshitz and Alfred van der Poorten, "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990), pp. 339-358.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Apéry's number or Apéry's constant zeta(3) is A002117. - N. J. A. Sloane, Jul 11 2023
Cf. A002736, A005258, A005429, A005430, A059415, A059416, A063007, A000172.
Cf. A006221, A000578, A006353.
Related to diagonal of rational functions: A268545-A268555.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.
Cf. A092826 (prime terms).

Programs

  • GAP
    List([0..20],n->Sum([0..n],k->Binomial(n,k)^2*Binomial(n+k,k)^2)); # Muniru A Asiru, Sep 28 2018
  • Haskell
    a005259 n = a005259_list !! n
a005259_list = 1 : 5 : zipWith div (zipWith (-)
   (tail $ zipWith (*) a006221_list a005259_list)
   (zipWith (*) (tail a000578_list) a005259_list)) (drop 2 a000578_list)
-- Reinhard Zumkeller, Mar 13 2014
  • Magma
    [&+[Binomial(n, k) ^2 *Binomial(n+k, k)^2: k in [0..n]]:n in  [0..17]]; // Marius A. Burtea, Jan 20 2020
  • Maple
    a := proc(n) option remember; if n=0 then 1 elif n=1 then 5 else (n^(-3))* ( (34*(n-1)^3 + 51*(n-1)^2 + 27*(n-1) +5)*a((n-1)) - (n-1)^3*a((n-1)-1)); fi; end;
# Alternative:
a := n -> hypergeom([-n, -n, 1+n, 1+n], [1, 1, 1], 1):
seq(simplify(a(n)), n=0..17); # Peter Luschny, Jan 19 2020
  • Mathematica
    Table[HypergeometricPFQ[{-n, -n, n+1, n+1}, {1,1,1}, 1],{n,0,13}] (* Jean-François Alcover, Apr 01 2011 *)
Table[Sum[(Binomial[n,k]Binomial[n+k,k])^2,{k,0,n}],{n,0,30}] (* Harvey P. Dale, Oct 15 2011 *)
a[ n_] := SeriesCoefficient[ SeriesCoefficient[ SeriesCoefficient[ SeriesCoefficient[ 1 / (1 - t (1 + x ) (1 + y ) (1 + z ) (x y z + (y + 1) (z + 1))), {t, 0, n}], {x, 0, n}], {y, 0, n}], {z, 0, n}]; (* Michael Somos, May 14 2016 *)
  • PARI
    a(n)=sum(k=0,n,(binomial(n,k)*binomial(n+k,k))^2) \\ Charles R Greathouse IV, Nov 20 2012
  • Python
    def A005259(n):
    m, g = 1, 0
    for k in range(n+1):
        g += m
        m *= ((n+k+1)*(n-k))**2
        m //=(k+1)**4
    return g # Chai Wah Wu, Oct 02 2022

Formula

D-finite with recurrence (n+1)^3*a(n+1) = (34*n^3 + 51*n^2 + 27*n + 5)*a(n) - n^3*a(n-1), n >= 1.
Representation as a special value of the hypergeometric function 4F3, in Maple notation: a(n)=hypergeom([n+1, n+1, -n, -n], [1, 1, 1], 1), n=0, 1, ... - Karol A. Penson Jul 24 2002
a(n) = Sum_{k >= 0} A063007(n, k)*A000172(k). A000172 = Franel numbers. - Philippe Deléham, Aug 14 2003
G.f.: (-1/2)*(3*x - 3 + (x^2-34*x+1)^(1/2))*(x+1)^(-2)*hypergeom([1/3,2/3],[1],(-1/2)*(x^2 - 7*x + 1)*(x+1)^(-3)*(x^2 - 34*x + 1)^(1/2)+(1/2)*(x^3 + 30*x^2 - 24*x + 1)*(x+1)^(-3))^2. - Mark van Hoeij, Oct 29 2011
Let g(x, y) = 4*cos(2*x) + 8*sin(y)*cos(x) + 5 and let P(n,z) denote the Legendre polynomial of degree n. Then G. A. Edgar posted a conjecture of Alexandru Lupas that a(n) equals the double integral 1/(4*Pi^2)*int {y = -Pi..Pi} int {x = -Pi..Pi} P(n,g(x,y)) dx dy. (Added Jan 07 2015: Answered affirmatively in Math Overflow question 178790) - Peter Bala, Mar 04 2012; edited by G. A. Edgar, Dec 10 2016
a(n) ~ (1+sqrt(2))^(4*n+2)/(2^(9/4)*Pi^(3/2)*n^(3/2)). - Vaclav Kotesovec, Nov 01 2012
a(n) = Sum_{k=0..n} C(n,k)^2 * C(n+k,k)^2. - Joerg Arndt, May 11 2013
0 = (-x^2+34*x^3-x^4)*y''' + (-3*x+153*x^2-6*x^3)*y'' + (-1+112*x-7*x^2)*y' + (5-x)*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
From Peter Bala, Jan 18 2020: (Start)
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j) * C(n,k)^2 * C(n+k,k)^2 * C(n,j) * C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} C(n,k) * C(n+k,k) * C(k,j)^3 (see Koepf, p. 55).
a(n) = Sum_{0 <= j, k <= n} C(n,k)^2 * C(n,j)^2 * C(3*n-j-k,2*n) (see Koepf, p. 119).
Diagonal coefficients of the rational function 1/((1 - x - y)*(1 - z - t) - x*y*z*t) (Straub, 2014). (End)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^m at m = 2. At m = 1 we get the Apéry numbers A005258. - Peter Bala, Dec 22 2020
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k). - Peter Bala, Jul 18 2024
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n,k)^2 * C(n,j)^2 * C(k+j,k), see Labelle et al. link. - Max Alekseyev, Mar 12 2025

A005258 Apéry numbers: a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(n+k,k).

Original entry on oeis.org

1, 3, 19, 147, 1251, 11253, 104959, 1004307, 9793891, 96918753, 970336269, 9807518757, 99912156111, 1024622952993, 10567623342519, 109527728400147, 1140076177397091, 11911997404064793, 124879633548031009, 1313106114867738897, 13844511065506477501
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
Equals the main diagonal of square array A108625. - Paul D. Hanna, Jun 14 2005
This sequence is t_5 in Cooper's paper. - Jason Kimberley, Nov 25 2012
Conjecture: For each n=1,2,3,... the polynomial a_n(x) = Sum_{k=0..n} C(n,k)^2*C(n+k,k)*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
Diagonal of rational functions 1/(1 - x - x*y - y*z - x*z - x*y*z), 1/(1 + y + z + x*y + y*z + x*z + x*y*z), 1/(1 - x - y - z + x*y + x*y*z), 1/(1 - x - y - z + y*z + x*z - x*y*z). - Gheorghe Coserea, Jul 07 2018

Examples

			G.f. = 1 + 3*x + 19*x^2 + 147*x^3 + 1251*x^4 + 11253*x^5 + 104959*x^6 + ...

References

  • Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
  • S. Melczer, An Invitation to Analytic Combinatorics, 2021; p. 129.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A002736, A005259, A005429, A005430, A108625, A143413, A218690, A218692.
Cf. A007318.
Cf. A001008, A002805.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.

Programs

  • GAP
    a:=n->Sum([0..n],k->(-1)^(n-k)*Binomial(n,k)*Binomial(n+k,k)^2);;
A005258:=List([0..20],n->a(n));; # Muniru A Asiru, Feb 11 2018
  • GAP
    List([0..20],n->Sum([0..n],k->Binomial(n,k)^2*Binomial(n+k,k))); # Muniru A Asiru, Jul 29 2018
  • Haskell
    a005258 n = sum [a007318 n k ^ 2 * a007318 (n + k) k | k <- [0..n]]
-- Reinhard Zumkeller, Jan 04 2013
  • Magma
    [&+[Binomial(n,k)^2 * Binomial(n+k,k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Nov 28 2018
  • Maple
    with(combinat): seq(add((multinomial(n+k,n-k,k,k))*binomial(n,k), k=0..n), n=0..18); # Zerinvary Lajos, Oct 18 2006
a := n -> binomial(2*n, n)*hypergeom([-n, -n, -n], [1, -2*n], 1):
seq(simplify(a(n)), n=0..20); # Peter Luschny, Feb 10 2018
  • Mathematica
    a[n_] := HypergeometricPFQ[ {n+1, -n, -n}, {1, 1}, 1]; Table[ a[n], {n, 0, 18}] (* Jean-François Alcover, Jan 20 2012, after Vladeta Jovovic *)
Table[Sum[Binomial[n,k]^2 Binomial[n+k,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Aug 25 2019 *)
  • PARI
    {a(n) = if( n<0, -(-1)^n * a(-1-n), sum(k=0, n, binomial(n, k)^2 * binomial(n+k, k)))} /* Michael Somos, Sep 18 2013 */
  • Python
    def A005258(n):
    m, g = 1, 0
    for k in range(n+1):
        g += m
        m *= (n+k+1)*(n-k)**2
        m //= (k+1)**3
    return g # Chai Wah Wu, Oct 02 2022

Formula

a(n) = hypergeom([n+1, -n, -n], [1, 1], 1). - Vladeta Jovovic, Apr 24 2003
D-finite with recurrence: (n+1)^2 * a(n+1) = (11*n^2+11*n+3) * a(n) + n^2 * a(n-1). - Matthijs Coster, Apr 28 2004
Let b(n) be the solution to the above recurrence with b(0) = 0, b(1) = 5. Then the b(n) are rational numbers with b(n)/a(n) -> zeta(2) very rapidly. The identity b(n)*a(n-1) - b(n-1)*a(n) = (-1)^(n-1)*5/n^2 leads to a series acceleration formula: zeta(2) = 5 * Sum_{n >= 1} 1/(n^2*a(n)*a(n-1)) = 5*(1/(1*3) + 1/(2^2*3*19) + 1/(3^2*19*147) + ...). Similar results hold for the constant e: see A143413. - Peter Bala, Aug 14 2008
G.f.: hypergeom([1/12, 5/12],[1], 1728*x^5*(1-11*x-x^2)/(1-12*x+14*x^2+12*x^3+x^4)^3) / (1-12*x+14*x^2+12*x^3+x^4)^(1/4). - Mark van Hoeij, Oct 25 2011
a(n) ~ ((11+5*sqrt(5))/2)^(n+1/2)/(2*Pi*5^(1/4)*n). - Vaclav Kotesovec, Oct 05 2012
1/Pi = 5*(sqrt(47)/7614)*Sum_{n>=0} (-1)^n a(n)*binomial(2n,n)*(682n+71)/15228^n. [Cooper, equation (4)] - Jason Kimberley, Nov 26 2012
a(-1 - n) = (-1)^n * a(n) if n>=0. a(-1 - n) = -(-1)^n * a(n) if n<0. - Michael Somos, Sep 18 2013
0 = a(n)*(a(n+1)*(+4*a(n+2) + 83*a(n+3) - 12*a(n+4)) + a(n+2)*(+32*a(n+2) + 902*a(n+3) - 147*a(n+4)) + a(n+3)*(-56*a(n+3) + 12*a(n+4))) + a(n+1)*(a(n+1)*(+17*a(n+2) + 374*a(n+3) - 56*a(n+4)) + a(n+2)*(+176*a(n+2) + 5324*a(n+3) - 902*a(n+4)) + a(n+3)*(-374*a(n+3) + 83*a(n+4))) + a(n+2)*(a(n+2)*(-5*a(n+2) - 176*a(n+3) + 32*a(n+4)) + a(n+3)*(+17*a(n+3) - 4*a(n+4))) for all n in Z. - Michael Somos, Aug 06 2016
a(n) = binomial(2*n, n)*hypergeom([-n, -n, -n],[1, -2*n], 1). - Peter Luschny, Feb 10 2018
a(n) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*binomial(n+k,k)^2. - Peter Bala, Feb 10 2018
G.f. y=A(x) satisfies: 0 = x*(x^2 + 11*x - 1)*y'' + (3*x^2 + 22*x - 1)*y' + (x + 3)*y. - Gheorghe Coserea, Jul 01 2018
From Peter Bala, Jan 15 2020: (Start)
a(n) = Sum_{0 <= j, k <= n} (-1)^(j+k)*C(n,k)*C(n+k,k)^2*C(n,j)* C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j)*C(n,k)^2*C(n+k,k)*C(n,j)* C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} (-1)^j*C(n,k)^2*C(n,j)*C(3*n-j-k,2*n). (End)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^m at m = 1. At m = 2 we get the Apéry numbers A005259. - Peter Bala, Dec 22 2020
a(n) = (-1)^n*Sum_{j=0..n} (1 - 5*j*H(j) + 5*j*H(n - j))*binomial(n, j)^5, where H(n) denotes the n-th harmonic number, A001008/A002805. (Paule/Schneider). - Peter Luschny, Jul 23 2021
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=(3 + x)*T(x) + (-1 + 22*x + 3*x^2)*T'(x) + x*(-1 + 11*x + x^2)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = 3*(1 - 12*x + 14*x^2 + 12*x^3 + x^4);
g3 = 1 - 18*x + 75*x^2 + 75*x^4 + 18*x^5 + x^6;
which determine an elliptic surface with four singular fibers. (End)
Conjecture: a(n)^2 = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A143007(n, k). - Peter Bala, Jul 08 2024

A005430 Apéry numbers: n*C(2*n,n).

Original entry on oeis.org

0, 2, 12, 60, 280, 1260, 5544, 24024, 102960, 437580, 1847560, 7759752, 32449872, 135207800, 561632400, 2326762800, 9617286240, 39671305740, 163352435400, 671560012200, 2756930576400, 11303415363240, 46290177201840, 189368906734800, 773942488394400
Offset: 0

Views

Author

Simon Plouffe

Keywords

Comments

Appears as diagonal in A003506. - Zerinvary Lajos, Apr 12 2006
The aerated sequence 1,0,2,0,12,0,60,0,... has e.g.f. 1+x*Bessel_I(1,2x). - Paul Barry, Mar 29 2010
Conjecture: the terms of the inverse binomial transform are 2*A132894(n). - R. J. Mathar, Oct 21 2012

References

  • Frank Harary and Edgar M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 78, (3.5.25).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A002011, A002457, A002736, A005258, A005259, A005429, 1/beta(n, n+1) in A061928.
Cf. A001803, A003506.

Programs

  • GAP
    List([0..30], n-> n*Binomial(2*n,n)); # G. C. Greubel, Dec 09 2018
  • Magma
    [n*Binomial(2*n,n): n in [0..30]]; // G. C. Greubel, Dec 09 2018
  • Maple
    A005430 := n -> n*binomial(2*n, n);
  • Mathematica
    Table[n*Binomial[2n,n],{n,0,30}] (* Harvey P. Dale, May 29 2015 *)
  • PARI
    a(n)=-(-1)^n*real(polcoeff(serlaplace(x^2*besselh1(1,2*x)),2*n)) \\ Ralf Stephan
  • Sage
    [n*binomial(2*n,n) for n in range(30)] # G. C. Greubel, Dec 09 2018

Formula

a(n) = A002011(n-1)/2 = 2 * A002457(n-1).
Sum_{n >= 1} 1/a(n) = Pi*sqrt(3)/9. - Benoit Cloitre, Apr 07 2002
G.f.: 2*x/sqrt((1-4*x)^3). - Marco A. Cisneros Guevara, Jul 25 2011
E.g.f.: a(n) = n!* [x^n] exp(2*x)*2*x*(BesselI(0, 2*x)+BesselI(1, 2*x)). - Peter Luschny, Aug 25 2012
D-finite with recurrence (-n+1)*a(n) + 2*(2*n-1)*a(n-1) = 0. - R. J. Mathar, Dec 03 2012
G.f.: 2*x*(1-4*x)^(-3/2) = -G(0)/2 where G(k) = 1 - (2*k+1)/(1 - 2*x/(2*x - (k+1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 06 2012
a(n-1) = Sum_{k=0..floor(n/2)} k*C(n,k)*C(n-k,k)*2^(n-2*k). - Robert FERREOL, Aug 29 2015
From Ilya Gutkovskiy, Jan 17 2017: (Start)
a(n) ~ 4^n*sqrt(n)/sqrt(Pi).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(phi)/sqrt(5) = A086466, where phi is the golden ratio. (End)
1/a(n) = (-1)^n*Sum_{j=0..n-1} binomial(n-1,j)*Bernoulli(j+n)/(j+n) for n >= 1. See the Amdeberhan & Cohen link. - Peter Luschny, Jun 20 2017
1/a(n) = Sum_{k=0..n} (-1)^(k+1)*binomial(n,k)*HarmonicNumber(n+k) for n >= 1. - Peter Luschny, Aug 15 2017
Sum_{n>=1} x^n/a(n) = 2*sqrt(x/(4-x))*arcsin(sqrt(x)/2), for abs(x) < 4 (Adegoke et al., 2022, section 6, p. 11). - Amiram Eldar, Dec 07 2024

Extensions

More terms from James Sellers, May 01 2000

A164102 Decimal expansion of 2*Pi^2.

Original entry on oeis.org

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

Views

Author

R. J. Mathar, Aug 10 2009

Keywords

Comments

Surface area of the 4-dimensional unit sphere. The volume of the 4-dimensional unit sphere is a fourth of this, A102753.
Also decimal expansion of Pi^2/5 = 1.973920..., with offset 1. - Omar E. Pol, Oct 04 2011

Examples

			19.739208802178717237668981...

References

  • L. A. Santalo, Integral Geometry and Geometric Probability, Addison-Wesley, 1976, see p. 15.

Links

Crossrefs

Cf. A000032, A002388, A002736, A005248, A091476, A013661.

Programs

Formula

Equals 2*A002388 = 4*A102753.
Pi^2/5 = Sum_{k>=1} Lucas(2*k)/(k^2*binomial(2*k,k)) = Sum_{k>=1} A005248(k)/A002736(k) (Seiffert, 1991). - Amiram Eldar, Jan 17 2022

A002544 a(n) = binomial(2*n+1,n)*(n+1)^2.

Original entry on oeis.org

1, 12, 90, 560, 3150, 16632, 84084, 411840, 1969110, 9237800, 42678636, 194699232, 878850700, 3931426800, 17450721000, 76938289920, 337206098790, 1470171918600, 6379820115900, 27569305764000, 118685861314020, 509191949220240, 2177742427450200, 9287309860732800
Offset: 0

Views

Author

N. J. A. Sloane, Simon Plouffe

Keywords

Comments

Coefficients for numerical differentiation.
Take the first n integers 1,2,3..n and find all combinations with repetitions allowed for the first n of them. Find the sum of each of these combinations to get this sequence. Example for 1 and 2: 1,2,1+1,1+2,2+2 gives sum of 12=a(2). - J. M. Bergot, Mar 08 2016
Let cos(x) = 1 -x^2/2 +x^4/4!-x^6/6!.. = Sum_i (-1)^i x^(2i)/(2i)! be the standard power series of the cosine, and y = 2*(1-cos(x)) = 4*sin^2(x/2) = x^2 -x^4/12 +x^6/360 ...= Sum_i 2*(-1)^(i+1) x^(2i)/(2i)! be a closely related series. Then this sequence represents the reversion x^2 = Sum_i 1/a(i) *y^(i+1). - R. J. Mathar, May 03 2022

References

  • C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 92.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A085373, A002457, A002674, A202543.
Equals A002736/2.
A diagonal of A331430.

Programs

  • Maple
    seq((n+1)^2*(binomial(2*n+2, n+1))/2, n=0..29); # Zerinvary Lajos, May 31 2006
  • Mathematica
    Table[Binomial[2n+1,n](n+1)^2,{n,0,20}] (* Harvey P. Dale, Mar 23 2011 *)
  • PARI
    a(n)=binomial(2*n+1,n)*(n+1)^2
  • PARI
    x='x+O('x^99); Vec((1+2*x)/(1-4*x)^(5/2)) \\ Altug Alkan, Jul 09 2016
  • Python
    from sympy import binomial
def a(n): return binomial(2*n + 1, n)*(n + 1)**2 # Indranil Ghosh, Apr 18 2017

Formula

G.f.: (1 + 2x)/(1 - 4x)^(5/2).
a(n-1) = sum(i_1 + i_2 + ... + i_n) where the sum is over 0 <= i_1 <= i_2 <= ... <= i_n <= n; a(n) = (n+1)^2 C(2n+1, n). - David Callan, Nov 20 2003
a(n) = (n+1)^2 * binomial(2*n+2,n+1)/2. - Zerinvary Lajos, May 31 2006
Asymptotics: a(n)-> (1/64) * (128*n^2+176*n+41) * 4^n * n^(-1/2)/(sqrt(Pi)), for n->infinity. - Karol A. Penson, Aug 05 2013
G.f.: 2F1(3/2,2;1;4x). - R. J. Mathar, Aug 09 2015
a(n) = A002457(n)*(n+1). - R. J. Mathar, Aug 09 2015
a(n) = A000217(n)*A000984(n). - J. M. Bergot, Mar 10 2016
a(n-1) = A001791(n)*n*(n+1)/2. - Anton Zakharov, Jul 04 2016
From Ilya Gutkovskiy, Jul 04 2016: (Start)
E.g.f.: ((1 + 2*x)*(1 + 8*x)*BesselI(0,2*x) + 2*x*(3 + 8*x)*BesselI(1,2*x))*exp(2*x).
Sum_{n>=0} 1/a(n) = Pi^2/9 = A100044. (End)
From Peter Bala, Apr 18 2017: (Start)
With x = y^2/(1 + y) we have log^2(1 + y) = Sum_{n >= 0} (-1)^n*x^(n+1)/a(n). See Shenton and Kemp.
Series reversion ( Sum_{n >= 0} (-1)^n*x^(n+1)/a(n) ) = Sum_{n >= 1} 2*x^n/(2*n)! = Sum_{n >= 1} x^n/A002674(n). (End)
D-finite with recurrence n^2*a(n) -2*(n+1)*(2*n+1)*a(n-1)=0. - R. J. Mathar, Feb 08 2021
Sum_{n>=0} (-1)^n/a(n) = 4*arcsinh(1/2)^2 = A202543^2. - Amiram Eldar, May 14 2022

A331431 Triangle read by rows: T(n,k) = (-1)^(n+k)*(n+k+1)*binomial(n,k)*binomial(n+k,k) for n >= k >= 0.

Original entry on oeis.org

1, -2, 6, 3, -24, 30, -4, 60, -180, 140, 5, -120, 630, -1120, 630, -6, 210, -1680, 5040, -6300, 2772, 7, -336, 3780, -16800, 34650, -33264, 12012, -8, 504, -7560, 46200, -138600, 216216, -168168, 51480, 9, -720, 13860, -110880, 450450, -1009008, 1261260, -823680, 218790
Offset: 0

Views

Author

N. J. A. Sloane, Jan 17 2020

Keywords

Comments

Tables I, III, IV on pages 92 and 93 of Ser have integer entries and are A331430, A331431 (the present sequence), and A331432.
Given the system of equations 1 = Sum_{j=0..n} H(i, j) * x(j) for i = 2..n+2 where H(i,j) = 1/(i+j-1) for 1 <= i,j <= n is the n X n Hilbert matrix, then the solutions are x(j) = T(n, j). - Michael Somos, Mar 20 2020 [Corrected by Petros Hadjicostas, Jul 09 2020]

Examples

			Triangle begins:
   1;
  -2,    6;
   3,  -24,    30;
  -4,   60,  -180,     140;
   5, -120,   630,   -1120,     630;
  -6,  210, -1680,    5040,   -6300,     2772;
   7, -336,  3780,  -16800,   34650,   -33264,   12012;
  -8,  504, -7560,   46200, -138600,   216216, -168168,   51480;
   9, -720, 13860, -110880,  450450, -1009008, 1261260, -823680, 218790;
  ...

References

  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 93. See Table III.

Links

Crossrefs

Columns 1 is A331433 or equally A007531, column 2 is A331434 or equally A054559; the last three diagonals are A002738, A002736, A002457.
Cf. A000290 (row sums), A002457,, A100071, A108666 (alternating row sums), A109188 (diagonal sums), A331322, A331323, A331430, A331432.

Programs

  • Magma
    [(-1)^(n+k)*(k+1)*(2*k+1)*Binomial(n+k+1,n-k)*Catalan(k): k in [0..n], n in [0..15]]; // G. C. Greubel, Mar 22 2022
  • Maple
    gf := k -> (1+x)^(-2*(k+1)): ser := k -> series(gf(k), x, 32):
T := (n, k) -> ((2*k+1)!/(k!)^2)*coeff(ser(k), x, n-k):
seq(seq(T(n,k), k=0..n),n=0..7); # Peter Luschny, Jan 18 2020
S:=(n,k)->(-1)^(n+k)*(n+k+1)!/((k!)^2*(n-k)!);
rho:=n->[seq(S(n,k),k=0..n)];
for n from 0 to 14 do lprint(rho(n)); od: # N. J. A. Sloane, Jan 18 2020
  • Mathematica
    Table[(-1)^(n+k)*(n+k+1)*Binomial[2*k,k]*Binomial[n+k,n-k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 22 2022 *)
  • Sage
    flatten([[(-1)^(n+k)*(2*k+1)*binomial(2*k,k)*binomial(n+k+1,n-k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Mar 22 2022

Formula

T(n, 0) = (-1)^n*A000027(n+1).
T(n, 1) = A331433(n-1) = (-1)^(n+1)*A007531(n+2).
T(n, 2) = A331434(n-2) = (-1)^n*A054559(n+3).
T(n, n-2) = A002738(n-2).
T(n, n-1) = (-1)*A002736(n).
T(n, n) = A002457(n).
T(2*n, n) = (-1)^n*(3*n+1)!/(n!)^3 = (-1)^n*A331322(n).
Sum_{k=0..n} T(n, k) = A000290(n+1) (row sums).
Sum_{k=0..n} (-1)^k*T(n, k) = (-1)^n*A108666(n+1) (alternating row sums).
Sum_{k=0..n} T(n-k, k) = (-1)^n*A109188(n+1) (diagonal sums).
2^n*Sum_{k=0..n} T(n, k)/2^k = (-1)^floor(n/2)*A100071(n+1) (positive half sums).
(-2)^n*Sum_{k=0..n} T(n, k)/(-2)^k = A331323(n) (negative half sums).
T(n, k) = ((2*k+1)!/(k!)^2)*[x^(n-k)] (1+x)^(-2*(k+1)). - Georg Fischer and Peter Luschny, Jan 18 2020
T(n,k) = (-1)^(n+k)*(n+k+1)!/((k!)^2*(n-k)!), for n >= k >= 0. - N. J. A. Sloane, Jan 18 2020
From Petros Hadjicostas, Jul 09 2020: (Start)
Michael Somos's formulas above can be restated as
Sum_{k=0..n} T(n,k)/(i+k) = 1 for i = 1..n+1.
These are special cases of the following formula that is alluded to (in some way) in Ser's book:
1 - Sum_{k=0..n} T(n,k)/(x + k) = (x-1)*...*(x-(n + 1))/(x*(x+1)*...*(x+n)).
Because T(n,k) = (-1)^(n+1)*(n + k + 1)*A331430(n,k) and Sum_{k=0..n} A331430(n,k) = (-1)^(n+1), one may derive this formula from Ser's second formula stated in A331430. (End)
T(2*n+1, n) = (-2)*(-27)^n*Pochhammer(4/3, n)*Pochhammer(5/3, n)/(n!*(n+1)!). - G. C. Greubel, Mar 22 2022

Extensions

Several typos in the data corrected by Georg Fischer and Peter Luschny, Jan 18 2020
Definition changed by N. J. A. Sloane, Jan 18 2020

A005429 Apéry numbers: n^3*C(2n,n).

Original entry on oeis.org

0, 2, 48, 540, 4480, 31500, 199584, 1177176, 6589440, 35443980, 184756000, 938929992, 4672781568, 22850118200, 110079950400, 523521630000, 2462025277440, 11465007358860, 52926189069600, 242433164404200, 1102772230560000, 4984806175188840, 22404445765690560
Offset: 0

Views

Author

Simon Plouffe

Keywords

References

  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.6.3.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A002736, A005258, A005259, A005429, A005430, A145438.

Programs

  • Magma
    [Binomial(2*n,n)*n^3 : n in [0..30]]; // Wesley Ivan Hurt, Oct 21 2014
  • Mathematica
    Table[n^3 Binomial[2n,n],{n,0,30}] (* Harvey P. Dale, Apr 08 2012 *)
CoefficientList[Series[(2*x*(2*x*(2*x+5)+1))/(1-4*x)^(7/2), {x,0,30}], x] (* Vincenzo Librandi, Oct 22 2014 *)
  • SageMath
    [n^3*binomial(2*n,n) for n in range(31)] # G. C. Greubel, Nov 19 2022

Formula

Sum_{n>=1} (-1)^(n+1) / a(n) = 2 * zeta(3) / 5.
G.f.: (2*x*(2*x*(2*x + 5) + 1))/(1-4*x)^(7/2). - Harvey P. Dale, Apr 08 2012
From Ilya Gutkovskiy, Jan 17 2017: (Start)
a(n) ~ 4^n*n^(5/2)/sqrt(Pi).
Sum_{n>=1} 1/a(n) = (1/2)*4F3(1,1,1,1; 3/2,2,2; 1/4) = A145438. (End)

Extensions

Entry revised by N. J. A. Sloane, Apr 06 2004

A328717 Decimal expansion of Sum_{k>=1} Kronecker(5,k)/k^2.

Original entry on oeis.org

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

Views

Author

Jianing Song, Nov 19 2019

Keywords

Comments

Let Chi() be a primitive character modulo d, the so-called Dirichlet L-series L(s,Chi) is the analytic continuation (see the functional equations involving L(s,Chi) in the MathWorld link entitled Dirichlet L-Series) of the sum Sum_{k>=1} Chi(k)/k^s, Re(s)>0 (if d = 1, the sum converges requires Re(s)>1).
If s != 1, we can represent L(s,Chi) in terms of the Hurwitz zeta function by L(s,Chi) = (Sum_{k=1..d} Chi(k)*zeta(s,k/d))/d^s.
L(s,Chi) can also be represented in terms of the polylog function by L(s,Chi) = (Sum_{k=1..d} Chi'(k)*polylog(s,u^k))/(Sum_{k=1..d} Chi'(k)*u^k), where Chi' is the complex conjugate of Chi, u is any primitive d-th root of unity.
If m is a positive integer, we have L(m,Chi) = (Sum_{k=1..d} Chi(k)*polygamma(m-1,k/d))/((-d)^m*(m-1)!).
In this sequence we have Chi = A080891 and s = 2.

Examples

			1 - 1/2^2 - 1/3^2 + 1/4^2 + 1/6^2 - 1/7^2 - 1/8^2 + 1/9^2 + ... = 4*Pi^2/(25*sqrt(5)) = 0.70621140325974096993100317576256402766024647185294...

Links

Crossrefs

Cf. A000045, A001906, A002736, A080891.
Decimal expansion of Sum_{k>=1} Kronecker(d,k)/k^2, where d is a fundamental discriminant: A309710 (d=-8), A103133 (d=-7), A006752 (d=-4), A086724 (d=-3), A013661 (d=1), this sequence (d=5), A328895 (d=8), A258414 (d=12).
Decimal expansion of Sum_{k>=1} Kronecker(5,k)/k^s: A086466 (s=1), this sequence (s=2), A328723 (s=3).

Programs

  • Mathematica
    RealDigits[4*Pi^2/(25*Sqrt[5]), 10, 102] // First
  • PARI
    default(realprecision, 100); 4*Pi^2/(25*sqrt(5))

Formula

Equals 4*Pi^2/(25*sqrt(5)).
Equals (zeta(2,1/5) - zeta(2,2/5) - zeta(2,3/5) + zeta(2,4/5))/25, where zeta(s,a) is the Hurwitz zeta function.
Equals (polylog(2,u) - polylog(2,u^2) - polylog(2,u^3) + polylog(2,u^4))/sqrt(5), where u = exp(2*Pi*i/5) is a 5th primitive root of unity, i = sqrt(-1).
Equals (polygamma(1,1/5) - polygamma(1,2/5) - polygamma(1,3/5) - polygamma(1,4/5))/25.
Equals Sum_{k>=1} Fibonacci(2*k)/(k^2*binomial(2*k,k)) = Sum_{k>=1} A001906(k)/A002736(k) (Seiffert, 1991). - Amiram Eldar, Jan 17 2022
Equals 1/(Product_{p prime == 1 or 4 (mod 5)} (1 - 1/p^2) * Product_{p prime == 2 or 3 (mod 5)} (1 + 1/p^2)). - Amiram Eldar, Dec 17 2023
Showing 1-10 of 16 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.