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 1-3 of 3 results.

A002421 Expansion of (1-4*x)^(3/2) in powers of x.

Original entry on oeis.org

1, -6, 6, 4, 6, 12, 28, 72, 198, 572, 1716, 5304, 16796, 54264, 178296, 594320, 2005830, 6843420, 23571780, 81880920, 286583220, 1009864680, 3580429320, 12765008880, 45741281820, 164668614552, 595340375688, 2160865067312, 7871722745208, 28772503827312
Offset: 0

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Author

Keywords

Comments

Terms that are not divisible by 12 have indices in A019469. - Ralf Stephan, Aug 26 2004
From Ralf Steiner, Apr 06 2017: (Start)
By analytic continuation to the entire complex plane there exist regularized values for divergent sums such as:
Sum_{k>=0} a(k)^2/8^k = 2F1(-3/2,-3/2,1,2).
Sum_{k>=0} a(k) / 2^k = -i. (End)

Examples

			G.f. = 1 - 6*x + 6*x^2 + 4*x^3 + 6*x^4 + 12*x^5 + 28*x^6 + 72*x^7 + 198*x^8 + 572*x^9 + ...
		

References

  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 55.
  • 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).
  • T. N. Thiele, Interpolationsrechnung. Teubner, Leipzig, 1909, p. 164.

Crossrefs

Programs

  • GAP
    Concatenation([1], List([1..40], n-> 12*Factorial(2*n-4) /( Factorial(n)*Factorial(n-2)) )) # G. C. Greubel, Jul 03 2019
  • Magma
    [1,-6] cat [12*Catalan(n-2)/n: n in [2..30]]; // Vincenzo Librandi, Jun 11 2012
    
  • Maple
    A002421 := n -> 3*4^(n-1)*GAMMA(-3/2+n)/(sqrt(Pi)*GAMMA(1+n)):
    seq(A002421(n), n=0..29); # Peter Luschny, Dec 14 2015
  • Mathematica
    CoefficientList[Series[(1-4x)^(3/2),{x,0,40}],x] (* Vincenzo Librandi, Jun 11 2012 *)
    a[n_]:= Binomial[ 3/2, n] (-4)^n; (* Michael Somos, Dec 04 2013 *)
    a[n_]:= SeriesCoefficient[(1-4x)^(3/2), {x, 0, n}]; (* Michael Somos, Dec 04 2013 *)
  • PARI
    {a(n) = binomial( 3/2, n) * (-4)^n}; /* Michael Somos, Dec 04 2013 */
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( (1 - 4*x + x * O(x^n))^(3/2), n))}; /* Michael Somos, Dec 04 2013 */
    
  • Sage
    ((1-4*x)^(3/2)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jul 03 2019
    

Formula

a(n) = Sum_{m=0..n} binomial(n, m)*K_m(4), where K_m(x) = K_m(n, 2, x) is a Krawtchouk polynomial. - Alexander Barg (abarg(AT)research.bell-labs.com)
a(n) ~ (3/4)*Pi^(-1/2)*n^(-5/2)*2^(2*n)*(1 + 15/8*n^-1 + ...). - Joe Keane (jgk(AT)jgk.org), Nov 22 2001
From Ralf Stephan, Mar 11 2004: (Start)
a(n) = 12*(2*n-4)! /(n!*(n-2)!), n > 1.
a(n) = 12*Cat(n-2)/n = 2(Cat(n-1) - 4*Cat(n-2)), in terms of Catalan numbers (A000108).
Terms that are not divisible by 12 have indices in A019469. (End)
Let rho(x)=(1/Pi)*(x*(4-x))^(3/2), then for n >= 4, a(n) = Integral_{x=0..4} (x^(n-4) *rho(x)) dx. - Groux Roland, Mar 16 2011
G.f.: (1-4*x)^(3/2) = 1 - 6*x + 12*x^2/(G(0) + 2*x); G(k) = (4*x+1)*k-2*x+2-2*x*(k+2)*(2*k+1)/G(k+1); for -1/4 <= x < 1/4, otherwise G(0) = 2*x; (continued fraction). - Sergei N. Gladkovskii, Dec 05 2011
G.f.: 1/G(0) where G(k) = 1 + 4*x*(2*k+1)/(1 - 1/(1 + (2*k+2)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 18 2012
G.f.: G(0)/2, where G(k) = 2 + 2*x*(2*k-3)*G(k+1)/(k+1). - Sergei N. Gladkovskii, Jun 06 2013 [Edited by Michael Somos, Dec 04 2013]
0 = a(n+2) * (a(n+1) - 14*a(n)) + a(n+1) * (6*a(n+1) + 16*a(n)) for all n in Z. - Michael Somos, Dec 04 2013
A232546(n) = 3^n * a(n). - Michael Somos, Dec 04 2013
G.f.: hypergeometric1F0(-3/2;;4*x). - R. J. Mathar, Aug 09 2015
a(n) = 3*4^(n-1)*Gamma(-3/2+n)/(sqrt(Pi)*Gamma(1+n)). - Peter Luschny, Dec 14 2015
From Ralf Steiner, Apr 06 2017: (Start)
Sum_{k>=0} a(k)/4^k = 0.
Sum_{k>=0} a(k)^2/16^k = 32/(3*Pi).
Sum_{k>=0} a(k)^2*(k/8)/16^k = 1/Pi.
Sum_{k>=0} a(k)^2*(-k/24+1/8)/16^k = 1/Pi.
Sum_{k>=0} a(k-1)^2*(k-1/4)/16^k = 1/Pi.
Sum_{k>=0} a(k-1)^2*(2k-2)/16^k = 1/Pi.(End)
D-finite with recurrence: n*a(n) +2*(-2*n+5)*a(n-1)=0. - R. J. Mathar, Feb 20 2020
From Amiram Eldar, Mar 22 2022: (Start)
Sum_{n>=0} 1/a(n) = 4/3 + 10*Pi/(81*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 92/75 - 4*sqrt(5)*log(phi)/125, where phi is the golden ratio (A001622). (End)

A096304 Numbers k such that 3k does not divide (6k-4)!/((3k-2)!*(3k-1)!).

Original entry on oeis.org

1, 2, 3, 4, 5, 9, 10, 11, 12, 13, 14, 27, 28, 29, 30, 31, 32, 36, 37, 38, 39, 40, 41, 81, 82, 83, 84, 85, 86, 90, 91, 92, 93, 94, 95, 108, 109, 110, 111, 112, 113, 117, 118, 119, 120, 121, 122, 243, 244, 245, 246, 247, 248, 252, 253, 254, 255, 256, 257, 270, 271
Offset: 1

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Author

Ralf Stephan, Aug 03 2004

Keywords

Comments

Equivalently, members of A019469 divisible by 3, divided by 3.
Ralf Stephan's formula is that terms k written in ternary have an arbitrary least significant digit and above that only 0's and 1's (per A340051). - Kevin Ryde, May 22 2021
{3a(n)-1:n>=1} is the set of positive integers k such that the k-th central binomial coefficient is not divisible by (k+1)*(2k-1). Such integers k are characterized by the following property: k is congruent to 2 (mod 3), and at least one of k-1, k+1 has no 2's in its base-3 expansion. - Valerio De Angelis, Aug 08 2022

Crossrefs

Cf. A340051 (ternary digits), A005836, A019469, A187358.

Programs

  • Mathematica
    Select[Range[300],Mod[(6#-4)!/((3#-2)!(3#-1)!),3#]!=0&] (* Harvey P. Dale, Jun 11 2019 *)
  • PARI
    for(n=1,300,if(((6*n-4)!/(3*n-2)!/(3*n-1)!)%(3*n),print1(n",")))
    
  • PARI
    a(n) = my(r);[n,r]=divrem(n,3); fromdigits(concat(binary(n),r), 3); \\ Kevin Ryde, May 22 2021
    
  • Python
    def A096304(n):
        a, b = divmod(n,3)
        return int(bin(a)[2:],3)*3+b # Chai Wah Wu, Jul 29 2025

Formula

a(n) = 9 * A005836(floor(n/6)) + (n mod 6) (conjectured) (confirmed, see links).
G.f.: x*(1+2*x)/(1-x^3) + 3/(1-x) * Sum_{i>=0} 3^i * x^(3*2^i) / (1 + x^(3*2^i)). - Kevin Ryde, May 22 2021

A019470 Numbers k that divide binomial(2*k-4, k-2).

Original entry on oeis.org

1, 5, 7, 10, 11, 13, 14, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 28, 29, 31, 34, 35, 37, 38, 40, 41, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79
Offset: 1

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Author

Keywords

Comments

Old name was: (n-2)-nd Catalan number is divisible by n.
For k in this sequence, convex k-gons have a number of triangulations that is divisible by k. - David Eppstein, Dec 24 2022

Crossrefs

Complement of A019469. Cf. A000108.

Programs

  • Maple
    A019470:=n->`if`(binomial(2*n-4,n-2) mod n = 0,n,NULL): seq(A019470(n), n=1..70); # Wesley Ivan Hurt, Sep 13 2014
  • Mathematica
    Select[Range[100],Divisible[Binomial[2#-4,#-2],#]&] (* Giorgos Kalogeropoulos, Aug 20 2021 *)
  • PARI
    valp(n,p)=my(s); while(n\=p, s+=n); s
    bin(n,p)=valp(2*n,p)-2*valp(n,p)
    is(n)=my(f=factor(n)); for(i=1,#f~, if(bin(n-2,f[i,1])Charles R Greathouse IV, Nov 04 2016

Extensions

More terms from Wesley Ivan Hurt, Sep 13 2014
Offset corrected by Robert Israel, Sep 14 2014
Name changed by Wesley Ivan Hurt, Sep 16 2014
Showing 1-3 of 3 results.