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 1-2 of 2 results.

A059366 Triangle T(m,s), m >= 0, 0 <= s <= m, arising in the computation of certain integrals.

Original entry on oeis.org

1, 1, 1, 3, 2, 3, 15, 9, 9, 15, 105, 60, 54, 60, 105, 945, 525, 450, 450, 525, 945, 10395, 5670, 4725, 4500, 4725, 5670, 10395, 135135, 72765, 59535, 55125, 55125, 59535, 72765, 135135, 2027025, 1081080, 873180, 793800, 771750, 793800, 873180
Offset: 0

Views

Author

N. J. A. Sloane, Jan 28 2001

Keywords

Comments

From Petros Hadjicostas, May 12 2020: (Start)
Following Comtet (1974, pp. 166-167), let J(m) = Integral_{t = 0..Pi/2} (A^2*cos^2(t) + B^2*sin^2(t))^(-m)) dt for m >= 0. Then J(m+1) = (Pi/(2^(m+1)*A*B*m!)) * Sum_{s=0..m} T(m,s)*A^(-2*s)*B^(-2*m+2*s).
Given m >= 0, the collection of numbers T(m,s)/A000165(m) = T(m,s)/(m!*2^m), s = 0..m, forms a discrete probability distribution on the set {0,1,...,m}, which is known as the "finite discrete arcsine distribution of order m". See Konrad (1969, Section 3.3) and Konrad (1992, Section 2.1, pp. 189-190). (End)

Examples

			Triangle T(m,s) (with rows m >= 0 and columns 0 <= s <= m) begins as follows:
    1;
    1,   1;
    3,   2,   3;
   15,   9,   9,  15;
  105,  60,  54,  60, 105;
  945, 525, 450, 450, 525, 945;
  ...
From _Petros Hadjicostas_, May 13 2020: (Start)
With m = 4, we have
J(4) = Integral_{t = 0..Pi/2} (A^2*cos^2(t) + B^2*sin^2(t))^(-4) dt
= Pi/(2^4*A*B*3!) * Sum_{s=0..3} T(3,s)*A^(-2*s)*B(-6+2*s)
= Pi/(96*A*B) * (15*B^(-6) + 9*A^(-2)*B^(-4) + 9*A^(-4)*B^(-2) + 15*A^(-6)). (End)
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 166-167; see a(m,s) (typo in a formula corrected below).

Crossrefs

Central diagonal gives A001757. Other diagonals and columns include A001147, A001193, A001194.

Programs

  • Magma
    /* as triangle */ [[Binomial(2*s,s)*Binomial(2*n-2*s, n-s)*Factorial(n)/2^n: s in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jan 09 2017
  • Maple
    A059366 = proc(m, s) option remember; if s = 0  then (2*m)!/(2^m*m!) else
    (2*s-1)*(m-s+1)/(s*(2*m-2*s+1)) * A059366(m, s-1) end if; end proc:
    seq(print(seq(A059366(m, s), s = 0..m)), m = 0..10) ; # Peter Bala, Apr 14 2024
  • Mathematica
    Table[Binomial[2*s, s]*Binomial[2*n - 2*s, n - s]*n!/2^n, {n, 0, 10}, {s, 0, n}] // Flatten (* G. C. Greubel, Jan 08 2017 *)
  • PARI
    for(n=0,10, for(s=0,n, print1(binomial(2*s, s)*binomial(2*n - 2*s, n - s)*n!/2^n, ", "))) \\ G. C. Greubel, Jan 08 2017
    

Formula

T(m+2, s) = (2*m+3)*(T(m+1, s-1) + T(m+1, s)) - 4*(m+1)^2*T(m, s-1).
T(m, s) = m!*Sum_{k=0..s} (-1)^k*2^(2*k-m)*binomial(s, k)*binomial(2*m-2*k, s)*binomial(2*m-2*k-s, m-k). [Typo in Comtet (1974, p. 166) corrected by Petros Hadjicostas, May 12 2020, using Comtet (1967, p. 85).]
From Reinhard Zumkeller, Apr 10 2004: (Start)
T(n,s) = A000984(s)*A000984(n-s)*A000142(n)/A000079(n).
T(n,s) = T(n,n-s).
Sum_{s=0..n} T(n,s) = A000165(n). (End)
From Petros Hadjicostas, May 13 2020: (Start)
T(m,s) = binomial(-1/2, s) * binomial(-1/2, m-s) * (-1)^m * m! * 2^m. [See Konrad (1992, pp. 189-190).]
T(m,m) = A001147(m) = T(m,0) for m >= 0.
T(m,m-1) = A001193(m-1) = T(m,1) for m >= 1.
T(m,m-2) = A001194(m) = T(m,2) for m >= 2.
T(m,m-3) = A001756(m) = T(m,3) for m >= 3.
T(m,floor(m/2)) = A001757(m) = T(m, ceiling(m/2)) for m >= 0.
Lim_{m -> infinity} Sum_{s: s/m <= x} T(m,s)/A000165(m) = (2/Pi)*arcsin(sqrt(x)) for x in [0,1], where the summation is over those s in {0,1,...,m} that satisfy s/m <= x. (End)
From Peter Bala, Apr 14 2024: (Start)
T(m, s) = (2*s - 1)*(m - s + 1)/(s*(2*m - 2*s + 1)) * T(m, s-1) for s >= 1.
T(m, s) = Sum_{i = 0..s} (-1)^(s-i)*binomial(m-i, s-i)*A368235(m, i). (End)

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Feb 08 2001

A283678 Number of possible draws of 2n pairs of consecutive cards from a set of 4n + 1 cards, so that the card that initially occupies the central position is not selected.

Original entry on oeis.org

1, 2, 54, 4500, 771750, 225042300, 99843767100, 62673358948200, 52880646612543750, 57733914846094987500, 79199384385873103852500, 133357363417740148141455000, 270426506783940730406180497500, 650063718230626755784087734375000, 1827886309419060919156885553671875000
Offset: 0

Views

Author

Keywords

Comments

The probability that the middle card is not selected in a random draw of 2n consecutive card pairs between 4n + 1 cards is a(n)/(4n)!!.
Essentially a bisection of A001757. - Giovanni Resta, Mar 14 2017

Examples

			For n = 1, you have 5 cards (A, B, C, D, E) and you can make 2 draws of pairs of consecutive cards (AB, DE) and (DE, AB) without select C.
		

Crossrefs

Programs

  • Maple
    ogf := sqrt(x) * BesselI(0, sqrt(x)/4) * BesselK(0,sqrt(x)/4) / 2;
    simplify(subs(x=1/x, asympt(ogf, x, 20))); # Mark van Hoeij, Oct 24 2017
  • Mathematica
    Table[Binomial[2n, n] Product[2n + 1 - 2i, {i, 1, n}]^2, {n, 0, 15}] (* Indranil Ghosh, Mar 22 2017 *)
  • PARI
    a(n)=binomial(2*n, n)*prod(i=1,n,2*n+1-2*i)^2 \\ Charles R Greathouse IV, Mar 14 2017
    
  • Python
    from sympy import binomial, factorial2
    print([binomial(2*n, n) * factorial2(2*n - 1)**2 for n in range(15)]) # Indranil Ghosh, Mar 22 2017

Formula

a(n) = binomial(2n, n)*((2n-1)!!)^2 = A092563(n)*A001147(n).
n*a(n) -2*(2*n-1)^3*a(n-1)=0. - R. J. Mathar, Jul 15 2017
a(n) ~ 2^(4*n+1)*(n/e)^(2*n)/sqrt(n*Pi). - Stefano Spezia, Mar 28 2025
Showing 1-2 of 2 results.