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.

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A274760 The multinomial transform of A001818(n) = ((2*n-1)!!)^2.

Original entry on oeis.org

1, 1, 10, 478, 68248, 21809656, 13107532816, 13244650672240, 20818058883902848, 48069880140604832128, 156044927762422185270016, 687740710497308621254625536, 4000181720339888446834235653120, 29991260979682976913756629498334208
Offset: 0

Views

Author

Johannes W. Meijer, Jul 27 2016

Keywords

Comments

The multinomial transform [MNL] transforms an input sequence b(n) into the output sequence a(n). Given the fact that the structure of the a(n) formulas, see the examples, lead to the multinomial coefficients A036039 the MNL transform seems to be an appropriate name for this transform. The multinomial transform is related to the exponential transform, see A274804 and the third formula. For the inverse multinomial transform [IML] see A274844.
To preserve the identity IML[MNL[b(n)]] = b(n) for n >= 0 for a sequence b(n) with offset 0 the shifted sequence b(n-1) with offset 1 has to be used as input for the MNL, otherwise information about b(0) will be lost in transformation.
In the a(n) formulas, see the examples, the multinomial coefficients A036039 appear.
We observe that a(0) = 1 and that this term provides no information about any value of b(n), this notwithstanding we will start the a(n) sequence with a(0) = 1.
The Maple programs can be used to generate the multinomial transform of a sequence. The first program uses the first formula which was found by Paul D. Hanna, see A158876, and Vladimir Kruchinin, see A215915. The second program uses properties of the e.g.f., see the sequences A158876, A213507, A244430 and A274539 and the third formula. The third program uses information about the inverse multinomial transform, see A274844.
Some MNL transform pairs are, n >= 1: A000045(n) and A244430(n-1); A000045(n+1) and A213527(n-1); A000108(n) and A213507(n-1); A000108(n-1) and A243953(n-1); A000142(n) and A158876(n-1); A000203(n) and A053529(n-1); A000110(n) and A274539(n-1); A000041(n) and A215915(n-1); A000035(n-1) and A177145(n-1); A179184(n) and A038205(n-1); A267936(n) and A000266(n-1); A267871(n) and A000090(n-1); A193356(n) and A088009(n-1).

Examples

			Some a(n) formulas, see A036039:
  a(0) = 1
  a(1) = 1*x(1)
  a(2) = 1*x(2) + 1*x(1)^2
  a(3) = 2*x(3) + 3*x(1)*x(2) + 1*x(1)^3
  a(4) = 6*x(4) + 8*x(1)*x(3) + 3*x(2)^2 + 6*x(1)^2*x(2) + 1*x(1)^4
  a(5) = 24*x(5) + 30*x(1)*x(4) + 20*x(2)*x(3) + 20*x(1)^2*x(3) + 15*x(1)*x(2)^2 + 10*x(1)^3*x(2) + 1*x(1)^5
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 1995, pp. 18-23.

Crossrefs

Programs

  • Maple
    nmax:= 13: b := proc(n): (doublefactorial(2*n-1))^2 end: a:= proc(n) option remember: if n=0 then 1 else add(((n-1)!/(n-k)!) * b(k) * a(n-k), k=1..n) fi: end: seq(a(n), n = 0..nmax); # End first MNL program.
    nmax:=13: b := proc(n): (doublefactorial(2*n-1))^2 end: t1 := exp(add(b(n)*x^n/n, n = 1..nmax+1)): t2 := series(t1, x, nmax+1): a := proc(n): n!*coeff(t2, x, n) end: seq(a(n), n = 0..nmax); # End second MNL program.
    nmax:=13: b := proc(n): (doublefactorial(2*n-1))^2 end: f := series(log(1+add(s(n)*x^n/n!, n=1..nmax)), x, nmax+1): d := proc(n): n*coeff(f, x, n) end: a(0) := 1: a(1) := b(1): s(1) := b(1): for n from 2 to nmax do s(n) := solve(d(n)-b(n), s(n)): a(n):=s(n): od: seq(a(n), n=0..nmax); # End third MNL program.
  • Mathematica
    b[n_] := (2*n - 1)!!^2;
    a[0] = 1; a[n_] := a[n] = Sum[((n-1)!/(n-k)!)*b[k]*a[n-k], {k, 1, n}];
    Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Nov 17 2017 *)

Formula

a(n) = Sum_{k=1..n} ((n-1)!/(n-k)!)*b(k)*a(n-k), n >= 1 and a(0) = 1, with b(n) = A001818(n) = ((2*n-1)!!)^2.
a(n) = n!*P(n), with P(n) = (1/n)*(Sum_{k=0..n-1} b(n-k)*P(k)), n >= 1 and P(0) = 1, with b(n) = A001818(n) = ((2*n-1)!!)^2.
E.g.f.: exp(Sum_{n >= 1} b(n)*x^n/n) with b(n) = A001818(n) = ((2*n-1)!!)^2.
denom(a(n)/2^n) = A001316(n); numer(a(n)/2^n) = [1, 1, 5, 239, 8531, 2726207, ...].

A274844 The inverse multinomial transform of A001818(n) = ((2*n-1)!!)^2.

Original entry on oeis.org

1, 8, 100, 1664, 34336, 843776, 24046912, 779780096, 28357004800, 1143189536768, 50612287301632, 2441525866790912, 127479926768287744, 7163315850315825152, 431046122080208896000, 27655699473265974050816, 1884658377677216933085184
Offset: 1

Views

Author

Johannes W. Meijer, Jul 27 2016

Keywords

Comments

The inverse multinomial transform [IML] transforms an input sequence b(n) into the output sequence a(n). The IML transform inverses the effect of the multinomial transform [MNL], see A274760, and is related to the logarithmic transform, see A274805 and the first formula.
To preserve the identity MNL[IML[b(n)]] = b(n) for n >= 0 for a sequence b(n) with offset 0 the shifted sequence b(n-1) with offset 1 has to be used as input for the MNL.
In the a(n) formulas, see the examples, the cumulant expansion numbers A127671 appear.
We observe that the inverse multinomial transform leaves the value of a(0) undefined.
The Maple programs can be used to generate the inverse multinomial transform of a sequence. The first program is derived from a formula given by Alois P. Heinz for the logarithmic transform, see the first formula and A001187. The second program uses the e.g.f. for multivariate row polynomials, see A127671 and the examples. The third program uses information about the inverse of the inverse of the multinomial transform, see A274760.
The IML transform of A001818(n) = ((2*n-1)!!)^2 leads quite unexpectedly to A005411(n), a sequence related to certain Feynman diagrams.
Some IML transform pairs, n >= 1: A000110(n) and 1/A000142(n-1); A137341(n) and A205543(n); A001044(n) and A003319(n+1); A005442(n) and A000204(n); A005443(n) and A001350(n); A007559(n) and A000244(n-1); A186685(n+1) and A131040(n-1); A061711(n) and A141151(n); A000246(n) and A000035(n); A001861(n) and A141044(n-1)/A001710(n-1); A002866(n) and A000225(n); A000262(n) and A000027(n).

Examples

			Some a(n) formulas, see A127671:
a(0) = undefined
a(1) = (1/0!) * (1*x(1))
a(2) = (1/1!) * (1*x(2) - x(1)^2)
a(3) = (1/2!) * (1*x(3) - 3*x(2)*x(1) + 2*x(1)^3)
a(4) = (1/3!) * (1*x(4) - 4*x(3)*x(1) - 3*x(2)^2 + 12*x(2)*x(1)^2 - 6*x(1)^4)
a(5) = (1/4!) * (1* x(5) - 5*x(4)*x(1) - 10*x(3)*x(2) + 20*x(3)*x(1)^2 + 30*x(2)^2*x(1) -60*x(2)*x(1)^3 + 24*x(1)^5)
		

References

  • Richard P. Feynman, QED, The strange theory of light and matter, 1985.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 1995, pp. 18-23.

Crossrefs

Programs

  • Maple
    nmax:=17: b := proc(n): (doublefactorial(2*n-1))^2 end: c:= proc(n) option remember; b(n) - add(k*binomial(n, k)*b(n-k)*c(k), k=1..n-1)/n end: a := proc(n): c(n)/(n-1)! end: seq(a(n), n=1..nmax); # End first IML program.
    nmax:=17: b := proc(n): (doublefactorial(2*n-1))^2 end: t1 := log(1+add(b(n)*x^n/n!, n=1..nmax+1)): t2 := series(t1, x, nmax+1): a := proc(n): n*coeff(t2, x, n) end: seq(a(n), n=1..nmax); # End second IML program.
    nmax:=17: b := proc(n): (doublefactorial(2*n-1))^2 end: f := series(exp(add(t(n)*x^n/n, n=1..nmax)), x, nmax+1): d := proc(n): n!*coeff(f, x, n) end: a(1):=b(1): t(1):= b(1): for n from 2 to nmax+1 do t(n) := solve(d(n)-b(n), t(n)): a(n):=t(n): od: seq(a(n), n=1..nmax); # End third IML program.
  • Mathematica
    nMax = 22; b[n_] := ((2*n-1)!!)^2; c[n_] := c[n] = b[n] - Sum[k*Binomial[n, k]*b[n-k]*c[k], {k, 1, n-1}]/n; a[n_] := c[n]/(n-1)!; Table[a[n], {n, 1, nMax}] (* Jean-François Alcover, Feb 27 2017, translated from Maple *)

Formula

a(n) = c(n)/(n-1)! with c(n) = b(n) - Sum_{k=1..n-1}(k*binomial(n, k)*b(n-k)*c(k)), n >= 1 and a(0) = undefined, with b(n) = A001818(n) = ((2*n-1)!!)^2.
a(n) = A000079(n-1) * A005411(n), n >= 1.

A111601 Exponential (binomial) convolution of A001818 (with interspersed zeros) and A000142 (factorials).

Original entry on oeis.org

1, 2, 9, 36, 225, 1350, 11025, 88200, 893025, 8930250, 108056025, 1296672300, 18261468225, 255660555150, 4108830350625, 65741285610000, 1187451971330625, 21374135483951250, 428670161650355625, 8573403233007112500
Offset: 1

Views

Author

Wolfdieter Lang, Aug 23 2005

Keywords

Crossrefs

Second column (m=1) of triangle |A111595|.

Formula

E.g.f.: (1/sqrt(1-x^2))*x/(1-x).
a(n) = n!*Sum_{k=0..floor((n-1)/2)} binomial(2*k, k)/(4^k).
a(2*k+1) = A111595(2*k+2, 0)= ((2*k+1)!!)^2 = A001818(k+1), k >= 0.
D-finite with recurrence (-n+1)*a(n) +n*a(n-1) +n*(n-1)^2*a(n-2)=0. a(1)=1, a(2)=2. - Sergei N. Gladkovskii, Jul 27 2012, corrected Aug 11 2025

A319150 a(n) = gcd(A275286(n), A001818(n+1)) / A025549(n+1)^2.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Tristan Cam, Nov 08 2018

Keywords

Comments

A(n) = a(n)*A025549(n+1)^2 = gcd(B(n), C(n)).
B(n) = A275286(n).
C(n) = A001818(n+1).
At first gcd(B(n), C(n)) = A025549(n+1)^2, but from n = 27 to n = 37, gcd(B(n), C(n)) = 11*A025549(n+1)^2, and then comes back to normal, then equals 19*A025549(n+1)^2, comes back to normal again, and so on ...
Let S(n) = Sum_{k=0..n} ((-1)^k)/(2*k+1)^2 (S(n) is NOT an integer sequence).
Notice that when n approaches +oo, D(n) converges to Catalan's constant (A006752).
A294970(n) is equal to the numerator of S(n) (when reduced).
Therefore B(n)/A(n) = A294970(n)
A294971(n) is equal to the denominator of S(n) (when reduced).
Therefore C(n)/A(n) = A294971(n).
This sequence was used to study the expression B(n)/C(n) (which equals S(n)) in an attempt to find out if Catalan's constant is irrational.

Examples

			For n = 5:
B(n) = A275286(5) = 98607816;
C(n) = A001818(5+1) = 108056025;
gcd(98607816,108056025) = A(5) = 9;
A025549(5+1)^2 = 3^2 = 9;
So a(5) = A(5)/A025549(5+1)^2 = 9/9 = 1.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := GCD[(2n+1)!!^2 * Sum[(-1)^k/(2k+1)^2, {k, 0, n}], (2n+1)!!^2]*
    LCM @@ Range[1, 2n+1, 2]^2 / ((2n+1)!!)^2; Array[a, 100, 0] (* Amiram Eldar, Nov 16 2018 *)
  • PARI
    dfo(n) = (2*n)! / n! / 2^n;
    a6(n) = dfo(n+1)^2*sum(k=0, n, (-1)^k/(2*k+1)^2);
    a8(n) = ((2*n)!/(n!*2^n))^2;
    a9(n) = (((2*n)!/n!)/2^n)/lcm(vector(n, i, 2*i-1));
    a(n) = gcd(a6(n) , a8(n+1)) / a9(n+1)^2; \\ Michel Marcus, Nov 08 2018

Formula

Explicit formula:
a(n) = gcd( ((2*n+1)!!)^2 * (Sum_{i=0..n}((-1)^i)/(2*i+1)^2), ((2*n+1)!!)^2 ) / ( (((2*n+1)!!)^2) / ( lcm{1,3,5,...,2*n+1} ) )^2.
A few relations:
gcd(A275286(n), A001818(n+1)) = a(n)*A025549(n+1)^2 = A(n);
A275286(n)/A(n) = A294970(n);
A001818(n+1)/A(n) = A294971(n);
Limit_{n->+oo} A294970(n)/A294971(n) = G (Catalan's Constant, decimal expansion: A006752).

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

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.

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).

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

A001147 Double factorial of odd numbers: a(n) = (2*n-1)!! = 1*3*5*...*(2*n-1).

Original entry on oeis.org

1, 1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425, 654729075, 13749310575, 316234143225, 7905853580625, 213458046676875, 6190283353629375, 191898783962510625, 6332659870762850625, 221643095476699771875, 8200794532637891559375, 319830986772877770815625
Offset: 0

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Comments

The solution to Schröder's third problem.
Number of fixed-point-free involutions in symmetric group S_{2n} (cf. A000085).
a(n-2) is the number of full Steiner topologies on n points with n-2 Steiner points. [corrected by Lyle Ramshaw, Jul 20 2022]
a(n) is also the number of perfect matchings in the complete graph K(2n). - Ola Veshta (olaveshta(AT)my-deja.com), Mar 25 2001
Number of ways to choose n disjoint pairs of items from 2*n items. - Ron Zeno (rzeno(AT)hotmail.com), Feb 06 2002
Number of ways to choose n-1 disjoint pairs of items from 2*n-1 items (one item remains unpaired). - Bartosz Zoltak, Oct 16 2012
For n >= 1 a(n) is the number of permutations in the symmetric group S_(2n) whose cycle decomposition is a product of n disjoint transpositions. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 21 2001
a(n) is the number of distinct products of n+1 variables with commutative, nonassociative multiplication. - Andrew Walters (awalters3(AT)yahoo.com), Jan 17 2004. For example, a(3)=15 because the product of the four variables w, x, y and z can be constructed in exactly 15 ways, assuming commutativity but not associativity: 1. w(x(yz)) 2. w(y(xz)) 3. w(z(xy)) 4. x(w(yz)) 5. x(y(wz)) 6. x(z(wy)) 7. y(w(xz)) 8. y(x(wz)) 9. y(z(wx)) 10. z(w(xy)) 11. z(x(wy)) 12. z(y(wx)) 13. (wx)(yz) 14. (wy)(xz) 15. (wz)(xy).
a(n) = E(X^(2n)), where X is a standard normal random variable (i.e., X is normal with mean = 0, variance = 1). So for instance a(3) = E(X^6) = 15, etc. See Abramowitz and Stegun or Hoel, Port and Stone. - Jerome Coleman, Apr 06 2004
Second Eulerian transform of 1,1,1,1,1,1,... The second Eulerian transform transforms a sequence s to a sequence t by the formula t(n) = Sum_{k=0..n} E(n,k)s(k), where E(n,k) is a second-order Eulerian number (A008517). - Ross La Haye, Feb 13 2005
Integral representation as n-th moment of a positive function on the positive axis: a(n) = Integral_{x=0..oo} x^n*exp(-x/2)/sqrt(2*Pi*x) dx, n >= 0. - Karol A. Penson, Oct 10 2005
a(n) is the number of binary total partitions of n+1 (each non-singleton block must be partitioned into exactly two blocks) or, equivalently, the number of unordered full binary trees with n+1 labeled leaves (Stanley, ex 5.2.6). - Mitch Harris, Aug 01 2006
a(n) is the Pfaffian of the skew-symmetric 2n X 2n matrix whose (i,j) entry is i for iDavid Callan, Sep 25 2006
a(n) is the number of increasing ordered rooted trees on n+1 vertices where "increasing" means the vertices are labeled 0,1,2,...,n so that each path from the root has increasing labels. Increasing unordered rooted trees are counted by the factorial numbers A000142. - David Callan, Oct 26 2006
Number of perfect multi Skolem-type sequences of order n. - Emeric Deutsch, Nov 24 2006
a(n) = total weight of all Dyck n-paths (A000108) when each path is weighted with the product of the heights of the terminal points of its upsteps. For example with n=3, the 5 Dyck 3-paths UUUDDD, UUDUDD, UUDDUD, UDUUDD, UDUDUD have weights 1*2*3=6, 1*2*2=4, 1*2*1=2, 1*1*2=2, 1*1*1=1 respectively and 6+4+2+2+1=15. Counting weights by height of last upstep yields A102625. - David Callan, Dec 29 2006
a(n) is the number of increasing ternary trees on n vertices. Increasing binary trees are counted by ordinary factorials (A000142) and increasing quaternary trees by triple factorials (A007559). - David Callan, Mar 30 2007
From Tom Copeland, Nov 13 2007, clarified in first and extended in second paragraph, Jun 12 2021: (Start)
a(n) has the e.g.f. (1-2x)^(-1/2) = 1 + x + 3*x^2/2! + ..., whose reciprocal is (1-2x)^(1/2) = 1 - x - x^2/2! - 3*x^3/3! - ... = b(0) - b(1)*x - b(2)*x^2/2! - ... with b(0) = 1 and b(n+1) = -a(n) otherwise. By the formalism of A133314, Sum_{k=0..n} binomial(n,k)*b(k)*a(n-k) = 0^n where 0^0 := 1. In this sense, the sequence a(n) is essentially self-inverse. See A132382 for an extension of this result. See A094638 for interpretations.
This sequence aerated has the e.g.f. e^(t^2/2) = 1 + t^2/2! + 3*t^4/4! + ... = c(0) + c(1)*t + c(2)*t^2/2! + ... and the reciprocal e^(-t^2/2); therefore, Sum_{k=0..n} cos(Pi k/2)*binomial(n,k)*c(k)*c(n-k) = 0^n; i.e., the aerated sequence is essentially self-inverse. Consequently, Sum_{k=0..n} (-1)^k*binomial(2n,2k)*a(k)*a(n-k) = 0^n. (End)
From Ross Drewe, Mar 16 2008: (Start)
This is also the number of ways of arranging the elements of n distinct pairs, assuming the order of elements is significant but the pairs are not distinguishable, i.e., arrangements which are the same after permutations of the labels are equivalent.
If this sequence and A000680 are denoted by a(n) and b(n) respectively, then a(n) = b(n)/n! where n! = the number of ways of permuting the pair labels.
For example, there are 90 ways of arranging the elements of 3 pairs [1 1], [2 2], [3 3] when the pairs are distinguishable: A = { [112233], [112323], ..., [332211] }.
By applying the 6 relabeling permutations to A, we can partition A into 90/6 = 15 subsets: B = { {[112233], [113322], [221133], [223311], [331122], [332211]}, {[112323], [113232], [221313], [223131], [331212], [332121]}, ....}
Each subset or equivalence class in B represents a unique pattern of pair relationships. For example, subset B1 above represents {3 disjoint pairs} and subset B2 represents {1 disjoint pair + 2 interleaved pairs}, with the order being significant (contrast A132101). (End)
A139541(n) = a(n) * a(2*n). - Reinhard Zumkeller, Apr 25 2008
a(n+1) = Sum_{j=0..n} A074060(n,j) * 2^j. - Tom Copeland, Sep 01 2008
From Emeric Deutsch, Jun 05 2009: (Start)
a(n) is the number of adjacent transpositions in all fixed-point-free involutions of {1,2,...,2n}. Example: a(2)=3 because in 2143=(12)(34), 3412=(13)(24), and 4321=(14)(23) we have 2 + 0 + 1 adjacent transpositions.
a(n) = Sum_{k>=0} k*A079267(n,k).
(End)
Hankel transform is A137592. - Paul Barry, Sep 18 2009
(1, 3, 15, 105, ...) = INVERT transform of A000698 starting (1, 2, 10, 74, ...). - Gary W. Adamson, Oct 21 2009
a(n) = (-1)^(n+1)*H(2*n,0), where H(n,x) is the probabilists' Hermite polynomial. The generating function for the probabilists' Hermite polynomials is as follows: exp(x*t-t^2/2) = Sum_{i>=0} H(i,x)*t^i/i!. - Leonid Bedratyuk, Oct 31 2009
The Hankel transform of a(n+1) is A168467. - Paul Barry, Dec 04 2009
Partial products of odd numbers. - Juri-Stepan Gerasimov, Oct 17 2010
See A094638 for connections to differential operators. - Tom Copeland, Sep 20 2011
a(n) is the number of subsets of {1,...,n^2} that contain exactly k elements from {1,...,k^2} for k=1,...,n. For example, a(3)=15 since there are 15 subsets of {1,2,...,9} that satisfy the conditions, namely, {1,2,5}, {1,2,6}, {1,2,7}, {1,2,8}, {1,2,9}, {1,3,5}, {1,3,6}, {1,3,7}, {1,3,8}, {1,3,9}, {1,4,5}, {1,4,6}, {1,4,7}, {1,4,8}, and {1,4,9}. - Dennis P. Walsh, Dec 02 2011
a(n) is the leading coefficient of the Bessel polynomial y_n(x) (cf. A001498). - Leonid Bedratyuk, Jun 01 2012
For n>0: a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = min(i,j)^2 for 1 <= i,j <= n. - Enrique Pérez Herrero, Jan 14 2013
a(n) is also the numerator of the mean value from 0 to Pi/2 of sin(x)^(2n). - Jean-François Alcover, Jun 13 2013
a(n) is the size of the Brauer monoid on 2n points (see A227545). - James Mitchell, Jul 28 2013
For n>1: a(n) is the numerator of M(n)/M(1) where the numbers M(i) have the property that M(n+1)/M(n) ~ n-1/2 (for example, large Kendell-Mann numbers, see A000140 or A181609, as n --> infinity). - Mikhail Gaichenkov, Jan 14 2014
a(n) = the number of upper-triangular matrix representations required for the symbolic representation of a first order central moment of the multivariate normal distribution of dimension 2(n-1), i.e., E[X_1*X_2...*X_(2n-2)|mu=0, Sigma]. See vignette for symmoments R package on CRAN and Phillips reference below. - Kem Phillips, Aug 10 2014
For n>1: a(n) is the number of Feynman diagrams of order 2n (number of internal vertices) for the vacuum polarization with one charged loop only, in quantum electrodynamics. - Robert Coquereaux, Sep 15 2014
Aerated with intervening zeros (1,0,1,0,3,...) = a(n) (cf. A123023), the e.g.f. is e^(t^2/2), so this is the base for the Appell sequence A099174 with e.g.f. e^(t^2/2) e^(x*t) = exp(P(.,x),t) = unsigned A066325(x,t), the probabilist's (or normalized) Hermite polynomials. P(n,x) = (a. + x)^n with (a.)^n = a_n and comprise the umbral compositional inverses for A066325(x,t) = exp(UP(.,x),t), i.e., UP(n,P(.,t)) = x^n = P(n,UP(.,t)), where UP(n,t) are the polynomials of A066325 and, e.g., (P(.,t))^n = P(n,t). - Tom Copeland, Nov 15 2014
a(n) = the number of relaxed compacted binary trees of right height at most one of size n. A relaxed compacted binary tree of size n is a directed acyclic graph consisting of a binary tree with n internal nodes, one leaf, and n pointers. It is constructed from a binary tree of size n, where the first leaf in a post-order traversal is kept and all other leaves are replaced by pointers. These links may point to any node that has already been visited by the post-order traversal. The right height is the maximal number of right-edges (or right children) on all paths from the root to any leaf after deleting all pointers. The number of unbounded relaxed compacted binary trees of size n is A082161(n). See the Genitrini et al. link. - Michael Wallner, Jun 20 2017
Also the number of distinct adjacency matrices in the n-ladder rung graph. - Eric W. Weisstein, Jul 22 2017
From Christopher J. Smyth, Jan 26 2018: (Start)
a(n) = the number of essentially different ways of writing a probability distribution taking n+1 values as a sum of products of binary probability distributions. See comment of Mitch Harris above. This is because each such way corresponds to a full binary tree with n+1 leaves, with the leaves labeled by the values. (This comment is due to Niko Brummer.)
Also the number of binary trees with root labeled by an (n+1)-set S, its n+1 leaves by the singleton subsets of S, and other nodes labeled by subsets T of S so that the two daughter nodes of the node labeled by T are labeled by the two parts of a 2-partition of T. This also follows from Mitch Harris' comment above, since the leaf labels determine the labels of the other vertices of the tree.
(End)
a(n) is the n-th moment of the chi-squared distribution with one degree of freedom (equivalent to Coleman's Apr 06 2004 comment). - Bryan R. Gillespie, Mar 07 2021
Let b(n) = 0 for n odd and b(2k) = a(k); i.e., let the sequence b(n) be an aerated version of this entry. After expanding the differential operator (x + D)^n and normal ordering the resulting terms, the integer coefficient of the term x^k D^m is n! b(n-k-m) / [(n-k-m)! k! m!] with 0 <= k,m <= n and (k+m) <= n. E.g., (x+D)^2 = x^2 + 2xD + D^2 + 1 with D = d/dx. The result generalizes to the raising (R) and lowering (L) operators of any Sheffer polynomial sequence by replacing x by R and D by L and follows from the disentangling relation e^{t(L+R)} = e^{t^2/2} e^{tR} e^{tL}. Consequently, these are also the coefficients of the reordered 2^n permutations of the binary symbols L and R under the condition LR = RL + 1. E.g., (L+R)^2 = LL + LR + RL + RR = LL + 2RL + RR + 1. (Cf. A344678.) - Tom Copeland, May 25 2021
From Tom Copeland, Jun 14 2021: (Start)
Lando and Zvonkin present several scenarios in which the double factorials occur in their role of enumerating perfect matchings (pairings) and as the nonzero moments of the Gaussian e^(x^2/2).
Speyer and Sturmfels (p. 6) state that the number of facets of the abstract simplicial complex known as the tropical Grassmannian G'''(2,n), the space of phylogenetic T_n trees (see A134991), or Whitehouse complex is a shifted double factorial.
These are also the unsigned coefficients of the x[2]^m terms in the partition polynomials of A134685 for compositional inversion of e.g.f.s, a refinement of A134991.
a(n)*2^n = A001813(n) and A001813(n)/(n+1)! = A000108(n), the Catalan numbers, the unsigned coefficients of the x[2]^m terms in the partition polynomials A133437 for compositional inversion of o.g.f.s, a refinement of A033282, A126216, and A086810. Then the double factorials inherit a multitude of analytic and combinatoric interpretations from those of the Catalan numbers, associahedra, and the noncrossing partitions of A134264 with the Catalan numbers as unsigned-row sums. (End)
Connections among the Catalan numbers A000108, the odd double factorials, values of the Riemann zeta function and its derivative for integer arguments, and series expansions of the reduced action for the simple harmonic oscillator and the arc length of the spiral of Archimedes are given in the MathOverflow post on the Riemann zeta function. - Tom Copeland, Oct 02 2021
b(n) = a(n) / (n! 2^n) = Sum_{k = 0..n} (-1)^n binomial(n,k) (-1)^k a(k) / (k! 2^k) = (1-b.)^n, umbrally; i.e., the normalized double factorial a(n) is self-inverse under the binomial transform. This can be proved by applying the Euler binomial transformation for o.g.f.s Sum_{n >= 0} (1-b.)^n x^n = (1/(1-x)) Sum_{n >= 0} b_n (x / (x-1))^n to the o.g.f. (1-x)^{-1/2} = Sum_{n >= 0} b_n x^n. Other proofs are suggested by the discussion in Watson on pages 104-5 of transformations of the Bessel functions of the first kind with b(n) = (-1)^n binomial(-1/2,n) = binomial(n-1/2,n) = (2n)! / (n! 2^n)^2. - Tom Copeland, Dec 10 2022

Examples

			a(3) = 1*3*5 = 15.
From _Joerg Arndt_, Sep 10 2013: (Start)
There are a(3)=15 involutions of 6 elements without fixed points:
  #:    permutation           transpositions
  01:  [ 1 0 3 2 5 4 ]      (0, 1) (2, 3) (4, 5)
  02:  [ 1 0 4 5 2 3 ]      (0, 1) (2, 4) (3, 5)
  03:  [ 1 0 5 4 3 2 ]      (0, 1) (2, 5) (3, 4)
  04:  [ 2 3 0 1 5 4 ]      (0, 2) (1, 3) (4, 5)
  05:  [ 2 4 0 5 1 3 ]      (0, 2) (1, 4) (3, 5)
  06:  [ 2 5 0 4 3 1 ]      (0, 2) (1, 5) (3, 4)
  07:  [ 3 2 1 0 5 4 ]      (0, 3) (1, 2) (4, 5)
  08:  [ 3 4 5 0 1 2 ]      (0, 3) (1, 4) (2, 5)
  09:  [ 3 5 4 0 2 1 ]      (0, 3) (1, 5) (2, 4)
  10:  [ 4 2 1 5 0 3 ]      (0, 4) (1, 2) (3, 5)
  11:  [ 4 3 5 1 0 2 ]      (0, 4) (1, 3) (2, 5)
  12:  [ 4 5 3 2 0 1 ]      (0, 4) (1, 5) (2, 3)
  13:  [ 5 2 1 4 3 0 ]      (0, 5) (1, 2) (3, 4)
  14:  [ 5 3 4 1 2 0 ]      (0, 5) (1, 3) (2, 4)
  15:  [ 5 4 3 2 1 0 ]      (0, 5) (1, 4) (2, 3)
(End)
G.f. = 1 + x + 3*x^2 + 15*x^3 + 105*x^4 + 945*x^5 + 10395*x^6 + 135135*x^7 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, (26.2.28).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 317.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 228, #19.
  • Hoel, Port and Stone, Introduction to Probability Theory, Section 7.3.
  • F. K. Hwang, D. S. Richards and P. Winter, The Steiner Tree Problem, North-Holland, 1992, see p. 14.
  • C. Itzykson and J.-B. Zuber, Quantum Field Theory, McGraw-Hill, 1980, pages 466-467.
  • 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).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 5.2.6 and also p. 178.
  • R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer-Verlag, New York, 1999, p. 73.
  • G. Watson, The Theory of Bessel Functions, Cambridge Univ. Press, 1922.

Crossrefs

Cf. A086677; A055142 (for this sequence, |a(n+1)| + 1 is the number of distinct products which can be formed using commutative, nonassociative multiplication and a nonempty subset of n given variables).
Constant terms of polynomials in A098503. First row of array A099020.
Subsequence of A248652.
Cf. A082161 (relaxed compacted binary trees of unbounded right height).
Cf. A053871 (binomial transform).

Programs

  • GAP
    A001147 := function(n) local i, s, t; t := 1; i := 0; Print(t, ", "); for i in [1 .. n] do t := t*(2*i-1); Print(t, ", "); od; end; A001147(100); # Stefano Spezia, Nov 13 2018
    
  • Haskell
    a001147 n = product [1, 3 .. 2 * n - 1]
    a001147_list = 1 : zipWith (*) [1, 3 ..] a001147_list
    -- Reinhard Zumkeller, Feb 15 2015, Dec 03 2011
    
  • Magma
    A001147:=func< n | n eq 0 select 1 else &*[ k: k in [1..2*n-1 by 2] ] >; [ A001147(n): n in [0..20] ]; // Klaus Brockhaus, Jun 22 2011
    
  • Magma
    I:=[1,3]; [1] cat [n le 2 select I[n]  else (3*n-2)*Self(n-1)-(n-1)*(2*n-3)*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 19 2015
    
  • Maple
    f := n->(2*n)!/(n!*2^n);
    A001147 := proc(n) doublefactorial(2*n-1); end: # R. J. Mathar, Jul 04 2009
    A001147 := n -> 2^n*pochhammer(1/2, n); # Peter Luschny, Aug 09 2009
    G(x):=(1-2*x)^(-1/2): f[0]:=G(x): for n from 1 to 29 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..19); # Zerinvary Lajos, Apr 03 2009; aligned with offset by Johannes W. Meijer, Aug 11 2009
    series(hypergeom([1,1/2],[],2*x),x=0,20); # Mark van Hoeij, Apr 07 2013
  • Mathematica
    Table[(2 n - 1)!!, {n, 0, 19}] (* Robert G. Wilson v, Oct 12 2005 *)
    a[ n_] := 2^n Gamma[n + 1/2] / Gamma[1/2]; (* Michael Somos, Sep 18 2014 *)
    Join[{1}, Range[1, 41, 2]!!] (* Harvey P. Dale, Jan 28 2017 *)
    a[ n_] := If[ n < 0, (-1)^n / a[-n], SeriesCoefficient[ Product[1 - (1 - x)^(2 k - 1), {k, n}], {x, 0, n}]]; (* Michael Somos, Jun 27 2017 *)
    (2 Range[0, 20] - 1)!! (* Eric W. Weisstein, Jul 22 2017 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(sum(binomial(n-1,i)*binomial(n-i-1,j)*a(i)*a(j)*a(n-i-j-1),j,0,n-i-1),i,0,n-1); /* Vladimir Kruchinin, May 06 2020 */
  • PARI
    {a(n) = if( n<0, (-1)^n / a(-n), (2*n)! / n! / 2^n)}; /* Michael Somos, Sep 18 2014 */
    
  • PARI
    x='x+O('x^33); Vec(serlaplace((1-2*x)^(-1/2))) \\ Joerg Arndt, Apr 24 2011
    
  • Python
    from sympy import factorial2
    def a(n): return factorial2(2 * n - 1)
    print([a(n) for n in range(101)])  # Indranil Ghosh, Jul 22 2017
    
  • Sage
    [rising_factorial(n+1,n)/2^n for n in (0..15)] # Peter Luschny, Jun 26 2012
    

Formula

E.g.f.: 1 / sqrt(1 - 2*x).
D-finite with recurrence: a(n) = a(n-1)*(2*n-1) = (2*n)!/(n!*2^n) = A010050(n)/A000165(n).
a(n) ~ sqrt(2) * 2^n * (n/e)^n.
Rational part of numerator of Gamma(n+1/2): a(n) * sqrt(Pi) / 2^n = Gamma(n+1/2). - Yuriy Brun, Ewa Dominowska (brun(AT)mit.edu), May 12 2001
With interpolated zeros, the sequence has e.g.f. exp(x^2/2). - Paul Barry, Jun 27 2003
The Ramanujan polynomial psi(n+1, n) has value a(n). - Ralf Stephan, Apr 16 2004
a(n) = Sum_{k=0..n} (-2)^(n-k)*A048994(n, k). - Philippe Deléham, Oct 29 2005
Log(1 + x + 3*x^2 + 15*x^3 + 105*x^4 + 945*x^5 + 10395*x^6 + ...) = x + 5/2*x^2 + 37/3*x^3 + 353/4*x^4 + 4081/5*x^5 + 55205/6*x^6 + ..., where [1, 5, 37, 353, 4081, 55205, ...] = A004208. - Philippe Deléham, Jun 20 2006
1/3 + 2/15 + 3/105 + ... = 1/2. [Jolley eq. 216]
Sum_{j=1..n} j/a(j+1) = (1 - 1/a(n+1))/2. [Jolley eq. 216]
1/1 + 1/3 + 2/15 + 6/105 + 24/945 + ... = Pi/2. - Gary W. Adamson, Dec 21 2006
a(n) = (1/sqrt(2*Pi))*Integral_{x>=0} x^n*exp(-x/2)/sqrt(x). - Paul Barry, Jan 28 2008
a(n) = A006882(2n-1). - R. J. Mathar, Jul 04 2009
G.f.: 1/(1-x-2x^2/(1-5x-12x^2/(1-9x-30x^2/(1-13x-56x^2/(1- ... (continued fraction). - Paul Barry, Sep 18 2009
a(n) = (-1)^n*subs({log(e)=1,x=0},coeff(simplify(series(e^(x*t-t^2/2),t,2*n+1)),t^(2*n))*(2*n)!). - Leonid Bedratyuk, Oct 31 2009
a(n) = 2^n*gamma(n+1/2)/gamma(1/2). - Jaume Oliver Lafont, Nov 09 2009
G.f.: 1/(1-x/(1-2x/(1-3x/(1-4x/(1-5x/(1- ...(continued fraction). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Dec 02 2009
The g.f. of a(n+1) is 1/(1-3x/(1-2x/(1-5x/(1-4x/(1-7x/(1-6x/(1-.... (continued fraction). - Paul Barry, Dec 04 2009
a(n) = Sum_{i=1..n} binomial(n,i)*a(i-1)*a(n-i). - Vladimir Shevelev, Sep 30 2010
E.g.f.: A(x) = 1 - sqrt(1-2*x) satisfies the differential equation A'(x) - A'(x)*A(x) - 1 = 0. - Vladimir Kruchinin, Jan 17 2011
a(n) = A123023(2*n). - Michael Somos, Jul 24 2011
a(n) = (1/2)*Sum_{i=1..n} binomial(n+1,i)*a(i-1)*a(n-i). See link above. - Dennis P. Walsh, Dec 02 2011
a(n) = Sum_{k=0..n} (-1)^k*binomial(2*n,n+k)*Stirling_1(n+k,k) [Kauers and Ko].
a(n) = A035342(n, 1), n >= 1 (first column of triangle).
a(n) = A001497(n, 0) = A001498(n, n), first column, resp. main diagonal, of Bessel triangle.
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term of M^n and sum of top row terms of M^(n-1), where M = a variant of the (1,2) Pascal triangle (Cf. A029635) as the following production matrix:
1, 2, 0, 0, 0, ...
1, 3, 2, 0, 0, ...
1, 4, 5, 2, 0, ...
1, 5, 9, 7, 2, ...
...
For example, a(3) = 15 is the left term in top row of M^3: (15, 46, 36, 8) and a(4) = 105 = (15 + 46 + 36 + 8).
(End)
G.f.: A(x) = 1 + x/(W(0) - x); W(k) = 1 + x + x*2*k - x*(2*k + 3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 17 2011
a(n) = Sum_{i=1..n} binomial(n,i-1)*a(i-1)*a(n-i). - Dennis P. Walsh, Dec 02 2011
a(n) = A009445(n) / A014481(n). - Reinhard Zumkeller, Dec 03 2011
a(n) = (-1)^n*Sum_{k=0..n} 2^(n-k)*s(n+1,k+1), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
a(n) = (2*n)4! = Gauss_factorial(2*n,4) = Product{j=1..2*n, gcd(j,4)=1} j. - Peter Luschny, Oct 01 2012
G.f.: (1 - 1/Q(0))/x where Q(k) = 1 - x*(2*k - 1)/(1 - x*(2*k + 2)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 19 2013
G.f.: 1 + x/Q(0), where Q(k) = 1 + (2*k - 1)*x - 2*x*(k + 1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 2*x*(2*k + 1)/(2*x*(2*k + 1) - 1 + 2*x*(2*k + 2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(2*k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: G(0), where G(k) = 1 + 2*x*(4*k + 1)/(4*k + 2 - 2*x*(2*k + 1)*(4*k + 3)/(x*(4*k + 3) + 2*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 22 2013
a(n) = (2*n - 3)*a(n-2) + (2*n - 2)*a(n-1), n > 1. - Ivan N. Ianakiev, Jul 08 2013
G.f.: G(0), where G(k) = 1 - x*(k+1)/(x*(k+1) - 1/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Aug 04 2013
a(n) = 2*a(n-1) + (2n-3)^2*a(n-2), a(0) = a(1) = 1. - Philippe Deléham, Oct 27 2013
G.f. of reciprocals: Sum_{n>=0} x^n/a(n) = 1F1(1; 1/2; x/2), confluent hypergeometric Function. - R. J. Mathar, Jul 25 2014
0 = a(n)*(+2*a(n+1) - a(n+2)) + a(n+1)*(+a(n+1)) for all n in Z. - Michael Somos, Sep 18 2014
a(n) = (-1)^n / a(-n) = 2*a(n-1) + a(n-1)^2 / a(n-2) for all n in Z. - Michael Somos, Sep 18 2014
From Peter Bala, Feb 18 2015: (Start)
Recurrence equation: a(n) = (3*n - 2)*a(n-1) - (n - 1)*(2*n - 3)*a(n-2) with a(1) = 1 and a(2) = 3.
The sequence b(n) = A087547(n), beginning [1, 4, 52, 608, 12624, ... ], satisfies the same second-order recurrence equation. This leads to the generalized continued fraction expansion lim_{n -> infinity} b(n)/a(n) = Pi/2 = 1 + 1/(3 - 6/(7 - 15/(10 - ... - n*(2*n - 1)/((3*n + 1) - ... )))). (End)
E.g.f of the sequence whose n-th element (n = 1,2,...) equals a(n-1) is 1-sqrt(1-2*x). - Stanislav Sykora, Jan 06 2017
Sum_{n >= 1} a(n)/(2*n-1)! = exp(1/2). - Daniel Suteu, Feb 06 2017
a(n) = A028338(n, 0), n >= 0. - Wolfdieter Lang, May 27 2017
a(n) = (Product_{k=0..n-2} binomial(2*(n-k),2))/n!. - Stefano Spezia, Nov 13 2018
a(n) = Sum_{i=0..n-1} Sum_{j=0..n-i-1} C(n-1,i)*C(n-i-1,j)*a(i)*a(j)*a(n-i-j-1), a(0)=1, - Vladimir Kruchinin, May 06 2020
From Amiram Eldar, Jun 29 2020: (Start)
Sum_{n>=1} 1/a(n) = sqrt(e*Pi/2)*erf(1/sqrt(2)), where erf is the error function.
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(Pi/(2*e))*erfi(1/sqrt(2)), where erfi is the imaginary error function. (End)
G.f. of reciprocals: R(x) = Sum_{n>=0} x^n/a(n) satisfies (1 + x)*R(x) = 1 + 2*x*R'(x). - Werner Schulte, Nov 04 2024

Extensions

Removed erroneous comments: neither the number of n X n binary matrices A such that A^2 = 0 nor the number of simple directed graphs on n vertices with no directed path of length two are counted by this sequence (for n = 3, both are 13). - Dan Drake, Jun 02 2009

A000246 Number of permutations in the symmetric group S_n that have odd order.

Original entry on oeis.org

1, 1, 1, 3, 9, 45, 225, 1575, 11025, 99225, 893025, 9823275, 108056025, 1404728325, 18261468225, 273922023375, 4108830350625, 69850115960625, 1187451971330625, 22561587455281875, 428670161650355625, 9002073394657468125, 189043541287806830625
Offset: 0

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Author

Keywords

Comments

Michael Reid (mreid(AT)math.umass.edu) points out that the e.g.f. for the number of permutations of odd order can be obtained from the cycle index for S_n, F(Y; X1, X2, X3, ... ) := e^(X1 Y + X2 Y^2/2 + X3 Y^3/3 + ... ) and is F(Y, 1, 0, 1, 0, 1, 0, ... ) = sqrt((1 + Y)/(1 - Y)).
a(n) appears to be the number of permutations on [n] whose up-down signature has nonnegative partial sums. For example, the up-down signature of (2,4,5,1,3) is (+1,+1,-1,+1) with nonnegative partial sums 1,2,1,2 and a(3)=3 counts (1,2,3), (1,3,2), (2,3,1). - David Callan, Jul 14 2006
This conjecture has been confirmed, see Bernardi, Duplantier, Nadeau link.
a(n) is the number of permutations of [n] for which all left-to-right minima occur in odd locations in the permutation. For example, a(3)=3 counts 123, 132, 231. Proof: For such a permutation of length 2n, you can append 1,2,..., or 2n+1 (2n+1 choices) and increase by 1 the original entries that weakly exceed the appended entry. This gives all such permutations of length 2n+1. But if the original length is 2n-1, you cannot append 1 (for then 1 would be a left-to-right min in an even location) so you can only append 2,3,..., or 2n (2n-1 choices). This count matches the given recurrence relation a(2n)=(2n-1)a(2n-1), a(2n+1)=(2n+1)a(2n). - David Callan, Jul 22 2008
a(n) is the n-th derivative of exp(arctanh(x)) at x = 0. - Michel Lagneau, May 11 2010
a(n) is the absolute value of the Moebius number of the odd partition poset on a set of n+1 points, where the odd partition poset is defined to be the subposet of the partition poset consisting of only partitions using odd part size (as well as the maximum element for n even). - Kenneth M Monks, May 06 2012
Number of permutations in S_n in which all cycles have odd length. - Michael Somos, Mar 17 2019
a(n) is the number of unranked labeled binary trees compatible with the binary labeled perfect phylogeny that, among possible two-leaf binary labeled perfect phylogenies for a sample of size n+2, is compatible with the smallest number of unranked labeled binary trees. - Noah A Rosenberg, Jan 16 2025

Examples

			For the Wallis numerators, denominators and partial products see A001900. - _Wolfdieter Lang_, Dec 06 2017
		

References

  • H.-D. Ebbinghaus et al., Numbers, Springer, 1990, p. 146.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 87.
  • 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).

Crossrefs

Bisections are A001818 and A079484.
Row sums of unsigned triangle A049218 and of A111594, A262125.
Main diagonal of A262124.
Cf. A002019.

Programs

  • Haskell
    a000246 n = a000246_list !! n
    a000246_list = 1 : 1 : zipWith (+)
       (tail a000246_list) (zipWith (*) a000246_list a002378_list)
    -- Reinhard Zumkeller, Feb 27 2012
    
  • Magma
    I:=[1,1]; [n le 2 select I[n] else Self(n-1)+(n^2-5*n+6)*Self(n-2): n in [1..30]]; // Vincenzo Librandi, May 02 2015
  • Maple
    a:= proc(n) option remember; `if`(n<2, 1,
          a(n-1) +(n-1)*(n-2)*a(n-2))
        end:
    seq(a(n), n=0..25);  # Alois P. Heinz, May 14 2018
  • Mathematica
    a[n_] := a[n] = a[n-1]*(n+Mod[n, 2]-1); a[0] = 1; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Nov 21 2011, after Pari *)
    a[n_] := a[n] = (n-2)*(n-3)*a[n-2] + a[n-1]; a[0] := 0; a[1] := 1; Table[a[i], {i, 0, 20}] (* or *)  RecurrenceTable[{a[0]==0, a[1]==1, a[n]==(n-2)*(n-3)a[n-2]+a[n-1]}, a, {n, 20}] (* G. C. Greubel, May 01 2015 *)
    CoefficientList[Series[Sqrt[(1+x)/(1-x)], {x, 0, 20}], x]*Table[k!, {k, 0, 20}] (* Stefano Spezia, Oct 07 2018 *)
  • PARI
    a(n)=if(n<1,!n,a(n-1)*(n+n%2-1))
    
  • PARI
    Vec( serlaplace( sqrt( (1+x)/(1-x) + O(x^55) ) ) )
    
  • PARI
    a(n)=prod(k=3,n,k+k%2-1) \\ Charles R Greathouse IV, May 01 2015
    
  • PARI
    a(n)=(n!/(n\2)!>>(n\2))^2/if(n%2,n,1) \\ Charles R Greathouse IV, May 01 2015
    

Formula

E.g.f.: sqrt(1-x^2)/(1-x) = sqrt((1+x)/(1-x)).
a(2*k) = (2*k-1)*a(2*k-1), a(2*k+1) = (2*k+1)*a(2*k), for k >= 0, with a(0) = 1.
Let b(1)=0, b(2)=1, b(k+2)=b(k+1)/k + b(k); then a(n+1) = n!*b(n+2). - Benoit Cloitre, Sep 03 2002
a(n) = Sum_{k=0..floor((n-1)/2)} (2k)! * C(n-1, 2k) * a(n-2k-1) for n > 0. - Noam Katz (noamkj(AT)hotmail.com), Feb 27 2001
Also successive denominators of Wallis's approximation to Pi/2 (unreduced): 1/1 * 2/1 * 2/3 * 4/3 * 4/5 * 6/5 * 6/7 * .., for n >= 1.
D-finite with recurrence: a(n) = a(n-1) + (n-1)*(n-2)*a(n-2). - Benoit Cloitre, Aug 30 2003
a(n) is asymptotic to (n-1)!*sqrt(2*n/Pi). - Benoit Cloitre, Jan 19 2004
a(n) = n! * binomial(n-1, floor((n-1)/2)) / 2^(n-1), n > 0. - Ralf Stephan, Mar 22 2004
E.g.f.: e^atanh(x), a(n) = n!*Sum_{m=1..n} Sum_{k=m..n} 2^(k-m)*Stirling1(k,m) *binomial(n-1,k-1)/k!, n > 0, a(0)=1. - Vladimir Kruchinin, Dec 12 2011
G.f.: G(0) where G(k) = 1 + x*(4*k-1)/((2*k+1)*(x-1) - x*(x-1)*(2*k+1)*(4*k+1)/(x*(4*k+1) + 2*(x-1)*(k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Jul 24 2012
G.f.: 1 + x*(G(0) - 1)/(x-1) where G(k) = 1 - (2*k+1)/(1-x/(x - 1/(1 - (2*k+1)/(1-x/(x - 1/G(k+1) ))))); (continued fraction). - Sergei N. Gladkovskii, Jan 15 2013
G.f.: G(0), where G(k) = 1 + x*(2*k+1)/(1 - x*(2*k+1)/(x*(2*k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 07 2013
For n >= 1, a(2*n) = (2*n-1)!!^2, a(2*n+1) = (2*n+1)*(2*n-1)!!^2. - Vladimir Shevelev, Dec 01 2013
E.g.f.: arcsin(x) - sqrt(1-x^2) + 1 for a(0) = 0, a(1) = a(2) = a(3) = 1. - G. C. Greubel, May 01 2015
Sum_{n>1} 1/a(n) = (L_0(1) + L_1(1))*Pi/2, where L is the modified Struve function. - Peter McNair, Mar 11 2022
From Peter Bala, Mar 29 2024: (Start)
a(n) = n! * Sum_{k = 0..n} (-1)^(n+k)*binomial(1/2, k)*binomial(-1/2, n-k).
a(n) = (1/4^n) * (2*n)!/n! * hypergeom([-1/2, -n], [1/2 - n], -1).
a(n) = n!/2^n * A063886(n). (End)

A053871 a(0)=1; a(1)=0; a(n) = 2*(n-1)*(a(n-1) + a(n-2)).

Original entry on oeis.org

1, 0, 2, 8, 60, 544, 6040, 79008, 1190672, 20314880, 387099936, 8148296320, 187778717632, 4702248334848, 127140703364480, 3691602647581184, 114562300528369920, 3784124901630435328, 132555364873399378432, 4908221631901073295360, 191549525877429961604096
Offset: 0

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Author

Cris Moore (moore(AT)santafe.edu) and Christian G. Bower, Mar 29 2000

Keywords

Comments

Number of deranged matchings of 2n people with partners (of either sex) other than their spouse. 2n objects are initially paired in some way and then are re-paired so that no object is with its original partner (the dancing problem in the article).
Of interest in the "collision problem", where, given a 2-to-1 function f, we are asked for x, y such that f(x)=f(y).
2^n*n!*a(n) = (2n)! b(n) where b(n) are the probabilities that appear in Margolis (2001). One interpretation is in terms of matchings or 1-factors of the complete graph on 2n vertices. The number of these is (2n)!/2^{n}n!. The number of 1-factors being disjoint from (that is, having no edges in common with) a given 1-factor is then a(n); and then b(n) is the probability of picking such a disjoint one factor at random.
Also n!*a(n) = 2^n * c(n) where c(n) = A001499(n). If we define d(n,k) = k(n-1)(d(n-1,k) + d(n-2,k)), with d(0,k) = 1 and d(1,k) = 0, so d(n,1) are the derangement numbers A000166, then a(n) = d(n,2) (cf. A033030, A088991). On the other hand, taking d*(n,k) = d(n,k)/k^{n}, we have d*(n,k) = (n-1)(d*(n-1,k) + d*(n-2,k)/k), with d*(0,k) = 1 and d*(1,k) = 0 and it is easy to see from Bricard's recurrence for c(n) that c(n)/n! satisfies this for k = 2.
A proof that the description in the first comment as "number of deranged matchings" implies the defining recursion relation: let (x_1, y_1), (x_2, y_2), ..., (x_n, y_n) be the given pairs. In a deranged matching x_1 will be paired with any of the 2(n-1) objects x_2, y_2, x_3, y_3, ..., x_n, y_n. It is sufficient to count only those deranged matchings in which x_1, is matched with x_2. They are of two kinds: (i) y_1 is not matched with y_2; their number is a(n-1); (ii) y_1 is matched with y_2; their number is a(n-2). - Emeric Deutsch, Jan 23 2009
Inverse binomial transform of the odd double factorials (A001147). - David Callan, Aug 25 2009
From Lewis Mammel (l_mammel(AT)att.net), Oct 07 2009: (Start)
The formula is given directly by the Principle of Inclusion and Exclusion.
The first term includes all pairings, the second term excludes all pairings containing each pair, the third term includes all pairings containing each pair of pairs, and so on.
Based on n-a -> n for large n, the ratio a(n)/(2n-1)!! -> exp(-1/2) ~= 0.60653.
We find a(n)/(2n-1)!! for n= 100, 200, 300, 400 ~= 0.6050124904, 0.6057720290, 0.6060250088, 0.6061514604. (End)
This is a subset of the set of pairings of the first 2n integers (A001147) in another way: forbidding pairs of the form (2k,2k+1) for all k as well as the pair (1,2n) (seeing the constraint as circular by opposition to the linear A165968). - Olivier Gérard, Feb 08 2011
a(n) is the n-th central moment of a central chi-squared distribution (with 1 degree of freedom), i.e., a(n) = E[ (Y- E[Y] )^n ] = E[ (X^2 - 1 )^n ] where Y is chi-squared, X is std normal, X~N(0,1), and the expectation operator is E[]. - David Fioramonti, May 11 2016
Exponential self-convolution of a(n)/2^n gives subfactorials (A000166). - Vladimir Reshetnikov, Oct 09 2016
For n > 1, also the number of maximum matchings in the n-cocktail party graph. - Eric W. Weisstein, Jun 15 2017
Number of polygonal tiles with n sides with two exits per side and n edges connecting pairs of exits, with no edges between exits on the same side (cf. A289191, A289269 for nonisomorphic tiles under rotational and rotational and reflectional symmetries). - Marko Riedel, Jun 28 2017
Number of ways n people can hold hands (every hand is holding another hand) where no one is holding their own hand. - Harry Richman, Aug 29 2023

References

  • R. Bricard, L'Intermédiaire des Mathématiciens, 8 (1901), 312-313.

Crossrefs

See A289191 for when rotational symmetries of the tiles are taken into account. - Marko Riedel, Jun 28 2017
Cf. A165968, number of pairings of 2n things disjoint to a given pairing, and containing a given pair not in the given pairing. It is given by a(n)/(2n-2). - Lewis Mammel (l_mammel(AT)att.net), Oct 07 2009

Programs

  • Haskell
    a053871 n = a053871_list !! n
    a053871_list = 1 : 0 : zipWith (*)
       [2,4..] (zipWith (+) a053871_list $ tail a053871_list)
    -- Reinhard Zumkeller, Mar 07 2012
  • Maple
    f:= gfun:-rectoproc({a(0) = 1, a(1) = 0, a(n) = 2*(n - 1)*(a(n - 1) + a(n - 2))},a(n),remember):
    map(f, [$0..30]); # Robert Israel, May 10 2016
  • Mathematica
    RecurrenceTable[{a[0]==1,a[1]==0,a[n]==2(n-1)(a[n-1]+a[n-2])}, a[n],{n,20}] (* Harvey P. Dale, Sep 15 2011 *)
    CoefficientList[Assuming[{Element[x, Reals], x>0}, Series[Sqrt[Pi/2] * (I + Erfi[Sqrt[(1+1/x)/2]]) / (E^((1+x)/(2*x)) * Sqrt[x*(x+1)]), {x, 0, 20}]], x] (* Vaclav Kotesovec, Feb 15 2015 *)
    Range[0, 20]! CoefficientList[Series[1/(Exp[x] Sqrt[1 - 2 x]), {x, 0, 20}], x] (* Eric W. Weisstein, Jun 15 2017 *)
    Table[(-1)^n HypergeometricPFQ[{1/2, -n}, {}, 2], {n, 20}] (* Eric W. Weisstein, Jun 15 2017 *)
    Table[I (-1)^n HypergeometricU[1/2, 3/2 + n, -1/2]/Sqrt[2], {n, 20}] (* Eric W. Weisstein, Dec 31 2017 *)
  • PARI
    a(n)=(-1)^(n+1)*sum(k=0,n,(-1)^k*binomial(n,k)*prod(i=0,k,2*i-1))
    

Formula

a(n) = A054479(n)/A001147(n).
E.g.f.: 1/(exp(x)*sqrt(1-2x)).
a(n) = (-1)^n*Sum_{k=0..n} (-1)^k*C(n, k)*(2*k-1)!!. - Benoit Cloitre, May 01 2003; corrected by David Fioramonti, May 17 2016
a(n) = Integral_{x>=0} (x-1)^n * (exp(-x/2)/sqrt(2*Pi*x)) dt. - Paul Barry, Apr 12 2010
Conjectured g.f.: T(0)/(1+x), where T(k) = 1 - x*(k+1)/(x*(k+1) - (1+x)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 13 2013
a(n) ~ 2^(n+1/2) * n^n / exp(n+1/2). - Vaclav Kotesovec, Mar 11 2014
G.f.: Sum_{k>=0} (2*k - 1)!!*x^k/(1 + x)^(k+1). - Ilya Gutkovskiy, Apr 12 2019
Conjecture: if m == n (mod q) for q odd, then (-1)^m*a(m) == (-1)^n*a(n) (mod q). - Harry Richman, Aug 29 2023

Extensions

More terms from James Sellers, Apr 08 2000

A008956 Triangle of central factorial numbers |4^k t(2n+1,2n+1-2k)| read by rows (n>=0, k=0..n).

Original entry on oeis.org

1, 1, 1, 1, 10, 9, 1, 35, 259, 225, 1, 84, 1974, 12916, 11025, 1, 165, 8778, 172810, 1057221, 893025, 1, 286, 28743, 1234948, 21967231, 128816766, 108056025, 1, 455, 77077, 6092515, 230673443, 3841278805, 21878089479, 18261468225, 1, 680
Offset: 0

Views

Author

Keywords

Comments

The n-th row gives the coefficients in the expansion of Product_{i=0..n-1} (x+(2i+1)^2), highest powers first (see the discussion of central factorial numbers in A008955). - N. J. A. Sloane, Feb 01 2011
Descending row polynomials in x^2 evaluated at k generate odd coefficients of e.g.f. sin(arcsin(kt)/k): 1, x^2 - 1, 9x^4 - 10x^2 + 1, 225x^6 - 259x^4 + 34x^2 - 1, ... - Ralf Stephan, Jan 16 2005
From Johannes W. Meijer, Jun 18 2009: (Start)
We define (Pi/2)*Beta(n-1/2-z/2,n-1/2+z/2)/Beta(n-1/2,n-1/2) = (Pi/2)*Gamma(n-1/2-z/2)* Gamma(n-1/2+z/2)/Gamma(n-1/2)^2 = sum(BG2[2m,n]*z^(2m), m = 0..infinity) with Beta(z,w) the Beta function. Our definition leads to BG2[2m,1] = 2*beta(2m+1) and the recurrence relation BG2[2m,n] = BG2[2m,n-1] - BG2[2m-2,n-1]/(2*n-3)^2 for m = -2, -1, 0, 1, 2, .. and n = 2, 3, .. , with beta(m) = sum((-1)^k/(1+2*k)^m, k=0..infinity). We observe that beta(2m+1) = 0 for m = -1, -2, -3, .. .We found for the BG2[2*m,n] = sum((-1)^(k+n)*t2(n-1,k-1)* 2*beta(2*m-2*n+2*k+1),k=1..n)/((2*n-3)!!)^2 with the central factorial numbers t2(n,m) as defined above; see also the Maple program.
From the BG2 matrix and the closely related EG2 and ZG2 matrices, see A008955, we arrive at the LG2 matrix which is defined by LG2[2m-1,1] = 2*lambda(2*m) and the recurrence relation LG2[2*m-1,n] = LG2[2*m-3,n-1]/((2*n-3)*(2*n-1)) - (2*n-3)*LG2[2*m-1,n-1]/(2*n-1) for m = -2, -1, 0, 1, 2, .. and n = 2, 3, .. , with lambda(m) = (1-2^(-m))*zeta(m) with zeta(m) the Riemann zeta function. We found for the matrix coefficients LG2[2m-1,n] = sum((-1)^(k+1)* t2(n-1,k-1)*2*lambda(2*m-2*n+2*k)/((2*n-1)!!*(2*n-3)!!), k=1..n) and we see that the central factorial numbers t2(n,m) once again play a crucial role.
(End)

Examples

			Triangle begins:
[1]
[1, 1]
[1, 10, 9]
[1, 35, 259, 225]
[1, 84, 1974, 12916, 11025]
[1, 165, 8778, 172810, 1057221, 893025]
[1, 286, 28743, 1234948, 21967231, 128816766, 108056025]
[1, 455, 77077, 6092515, 230673443, 3841278805, 21878089479, 18261468225]
...
		

References

  • P. L. Butzer, M. Schmidt, E. L. Stark and L. Vogt, Central Factorial Numbers: Their main properties and some applications, Numerical Functional Analysis and Optimization, 10 (5&6), 419-488 (1989). [From Johannes W. Meijer, Jun 18 2009]
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.

Crossrefs

Cf. A008958.
Columns include A000447, A001823. Right-hand columns include A001818, A001824, A001825. Cf. A008955.
Appears in A160480 (Beta triangle), A160487 (Lambda triangle), A160479 (ZL(n) sequence), A161736, A002197 and A002198. - Johannes W. Meijer, Jun 18 2009
Cf. A162443 (BG1 matrix) and A162448 (LG1 matrix). - Johannes W. Meijer, Jul 06 2009
Cf. A001147.

Programs

  • Haskell
    a008956 n k = a008956_tabl !! n !! k
    a008956_row n = a008956_tabl !! n
    a008956_tabl = [1] : f [1] 1 1 where
       f xs u t = ys : f ys v (t * v) where
         ys = zipWith (+) (xs ++ [t^2]) ([0] ++ map (* u^2) (init xs) ++ [0])
         v = u + 2
    -- Reinhard Zumkeller, Dec 24 2013
  • Maple
    f:=n->mul(x+(2*i+1)^2,i=0..n-1);
    for n from 0 to 12 do
    t1:=eval(f(n)); t1d:=degree(t1);
    t12:=y^t1d*subs(x=1/y,t1); t2:=seriestolist(series(t12,y,20));
    lprint(t2);
    od: # N. J. A. Sloane, Feb 01 2011
    A008956 := proc(n,k) local i ; mul( x+2*i-2*n-1,i=1..2*n) ; expand(%) ; coeftayl(%,x=0,2*(n-k)) ; abs(%) ; end: for n from 0 to 10 do for k from 0 to n do printf("%a,",A008956(n,k)) ; od: od: # R. J. Mathar, May 29 2009
    nmax:=7: for n from 0 to nmax do t2(n, 0):=1: t2(n, n):=(doublefactorial(2*n-1))^2 od: for n from 1 to nmax do for k from 1 to n-1 do t2(n, k) := (2*n-1)^2*t2(n-1, k-1)+t2(n-1, k) od: od: seq(seq(t2(n, k), k=0..n), n=0..nmax); # Johannes W. Meijer, Jun 18 2009, Revised Sep 16 2012
  • Mathematica
    t[, 0] = 1; t[n, n_] := t[n, n] = ((2*n-1)!!)^2; t[n_, k_] := t[n, k] = (2*n-1)^2*t[n-1, k-1] + t[n-1, k]; Table[t[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 07 2014, after Johannes W. Meijer *)
  • PARI
    {T(n, k) = if( n<=0, k==0, (-1)^k * polcoeff( numerator( 2^(2*n -1) / sum(j=0, 2*n - 1, binomial( 2*n - 1, j) / (x + 2*n - 1 - 2*j))), 2*n - 2*k))}; /* Michael Somos, Feb 24 2003 */
    

Formula

Conjecture row sums: Sum_{k=0..n} T(n,k) = |A101927(n+1)|. - R. J. Mathar, May 29 2009
May be generated by the recurrence t2(n,k) = (2*n-1)^2*t2(n-1,k-1)+t2(n-1,k) with t2(n,0) = 1 and t2(n,n)=((2*n-1)!!)^2. - Johannes W. Meijer, Jun 18 2009

Extensions

More terms from Vladeta Jovovic, Apr 16 2000
Edited by N. J. A. Sloane, Feb 01 2011

A079484 a(n) = (2n-1)!! * (2n+1)!!, where the double factorial is A006882.

Original entry on oeis.org

1, 3, 45, 1575, 99225, 9823275, 1404728325, 273922023375, 69850115960625, 22561587455281875, 9002073394657468125, 4348001449619557104375, 2500100833531245335015625, 1687568062633590601135546875, 1321365793042101440689133203125
Offset: 0

Views

Author

Benoit Cloitre, Jan 17 2003

Keywords

Comments

a(n) is the determinant of M(2n+1) where M(k) is the k X k matrix with m(i,j)=j if i+j=k m(i,j)=i otherwise. - Adapted to offset 0, Rainer Rosenthal, Jun 19 2024
In the following two comments on the calculation of the terms using permanents, offset 1 is assumed. In the corresponding PARI code, this is implemented with offset 0. - Hugo Pfoertner, Jun 23 2024
(-1)^n*a(n)/2^(2n-1) is the permanent of the (m X m) matrix {1/(x_i-y_j), 1<=i<=m, 1<=j<=m}, where x_1,x_2,...,x_m are the zeros of x^m-1 and y_1,y_2,...,y_m the zeros of y^m+1 and m=2n-1.
In 1881, R. F. Scott posed a conjecture that the absolute value of permanent of square matrix with elements a(i,j)= (x_i - y_j)^(-1), where x_1,...,x_n are roots of x^n=1, while y_1,...,y_n are roots of y^n=-1, equals a((n-1)/2)/2^n, if n>=1 is odd, and 0, if n>=2 is even. After a century (in 1979), the conjecture was proved by H. Minc. - Vladimir Shevelev, Dec 01 2013
a(n) is the number of permutations in S_{2n+1} in which all cycles have odd length. - José H. Nieto S., Jan 09 2012
Number of 3-bundled increasing bilabeled trees with 2n labels. - Markus Kuba, Nov 18 2014
a(n) is the number of rooted, binary, leaf-labeled topologies with 2n+2 leaves that have n+1 cherry nodes. - Noah A Rosenberg, Feb 12 2019

Examples

			G.f. = 1 + 3*x + 45*x^2 + 1575*x^3 + 99225*x^4 + 9823275*x^5 + ...
M(5) =
  [1, 2, 3, 1, 5]
  [1, 2, 2, 4, 5]
  [1, 3, 3, 4, 5]
  [4, 2, 3, 4, 5]
  [1, 2, 3, 4, 5].
Integral_{x=0..oo} x^3*BesselK(1, sqrt(x)) = 1575*Pi. - _Olivier Gérard_, May 20 2009
		

References

  • Miklós Bóna, A walk through combinatorics, World Scientific, 2006.

Crossrefs

Bisection of A000246, A053195, |A013069|, |A046126|. Cf. A000909.
Cf. A001044, A010791, |A129464|, A114779, are also values of similar moments.
Equals the row sums of A162005.
Cf. A316728.
Diagonal elements of A306364 in even-numbered rows.

Programs

  • Magma
    I:=[1, 3]; [n le 2 select I[n] else (4*n^2-8*n+3)*Self(n-1): n in [1..20]]; // Vincenzo Librandi, Nov 18 2014
    
  • Maple
    a:= n-> (d-> d(2*n-1)*d(2*n+1))(doublefactorial):
    seq(a(n), n=0..15);  # Alois P. Heinz, Jan 30 2013
    # second Maple program:
    A079484 := n-> LinearAlgebra[Determinant](Matrix(2*n+1, (i, j)-> `if`(i+j=2*n+1, j, i))): seq(A079484(n), n=0..14); # Rainer Rosenthal, Jun 18 2024
  • Mathematica
    a[n_] := (2n - 1)!!*(2n + 1)!!; Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Jan 30 2013 *)
  • PARI
    /* Formula using the zeta function and a log integral:*/
    L(n)= intnum(t=0, 1, log(1-1/t)^n);
    Zetai(n)= -I*I^n*(2*Pi)^(n-1)/(n-1)*L(1-n);
    a(m)={my(n=m+1);round(real(-I*2^(2*n-1)*Zetai(1/2-n)*L(-1/2+n)/(Zetai(-1/2+n)*L(1/2-n))))};
    /* Gerry Martens, Mar 07 2011, adapted to offset 0 by Hugo Pfoertner, Jun 19 2024 */
    
  • PARI
    {a(n) = if( n<0, -1 / self()(-1-n), (2*n + 1)! * (2*n)! / (n! * 2^n)^2 )}; /* Michael Somos, May 04 2017 */
    
  • PARI
    {a(n) = if( n<0, -1 / self()(-1-n), my(m = 2*n + 1); m! * polcoeff( x / sqrt( 1 - x^2 + x * O(x^m) ), m))}; /* Michael Somos, May 04 2017 */
    
  • PARI
    \\ using the Pochhammer symbol
    a(n) = {my(P(x,k)=gamma(x+k)/gamma(x)); 4^n*round(P(1/2,n)*P(3/2,n))} \\ Hugo Pfoertner, Jun 20 2024
    
  • PARI
    \\ Scott's (1881) method
    a(n) = {my(m=2*n+1, X = polroots(x^m-1), Y = polroots(x^m+1), M = matrix(m, m, i, j, 1/(X[i]-Y[j]))); (-1)^n * round(2^m * real(matpermanent(M)))}; \\ Hugo Pfoertner, Jun 23 2024

Formula

D-finite with recurrence a(n) = (4*n^2 - 1) * a(n-1) for all n in Z.
a(n) = A001147(n)*A001147(n+1).
E.g.f.: 1/(1-x^2)^(3/2) (with interpolated zeros). - Paul Barry, May 26 2003
a(n) = (2n+1)! * C(2n, n) / 2^(2n). - Ralf Stephan, Mar 22 2004.
Alternatingly signed values have e.g.f. sqrt(1+x^2).
a(n) is the value of the n-th moment of (1/Pi)*BesselK(1, sqrt(x)) on the positive part of the real line. - Olivier Gérard, May 20 2009
a(n) = -2^(2*n-1)*exp(i*n*Pi)*gamma(1/2+n)/gamma(3/2-n). - Gerry Martens, Mar 07 2011
E.g.f. (odd powers) tan(arcsin(x)) = Sum_{n>=0} (2n-1)!!*(2n+1)!!*x^(2*n+1)/(2*n+1)!. - Vladimir Kruchinin, Apr 22 2011
G.f.: 1 + x*(G(0) - 1)/(x-1) where G(k) = 1 - ((2*k+2)^2-1)/(1-x/(x - 1/G(k+1))); ( continued fraction ). - Sergei N. Gladkovskii, Jan 15 2013
a(n) = (2^(2*n+3)*Gamma(n+3/2)*Gamma(n+5/2))/Pi. - Jean-François Alcover, Jul 20 2015
Limit_{n->oo} 4^n*(n!)^2/a(n) = Pi/2. - Daniel Suteu, Feb 05 2017
From Michael Somos, May 04 2017: (Start)
a(n) = (2*n + 1) * A001818(n).
E.g.f.: Sum_{n>=0} a(n) * x^(2*n+1) / (2*n+1)! = x / sqrt(1 - x^2) = tan(arcsin(x)).
Given e.g.f. A(x) = y, then x * y' = y + y^3.
a(n) = -1 / a(-1-n) for all n in Z.
0 = +a(n)*(+288*a(n+2) -60*a(n+3) +a(n+4)) +a(n+1)*(-36*a(n+2) -4*a(n+3)) +a(n+2)*(+3*a(n+2)) for all n in Z. (End)
a(n) = Sum_{k=0..2n} (k+1) * A316728(n,k). - Alois P. Heinz, Jul 12 2018
From Amiram Eldar, Mar 18 2022: (Start)
Sum_{n>=0} 1/a(n) = 1 + L_1(1)*Pi/2, where L is the modified Struve function.
Sum_{n>=0} (-1)^n/a(n) = 1 - H_1(1)*Pi/2, where H is the Struve function. (End)

Extensions

Simpler description from Daniel Flath (deflath(AT)yahoo.com), Mar 05 2004
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