A057977 GCD of consecutive central binomial coefficients: a(n) = gcd(A001405(n+1), A001405(n)).
1, 1, 1, 3, 2, 10, 5, 35, 14, 126, 42, 462, 132, 1716, 429, 6435, 1430, 24310, 4862, 92378, 16796, 352716, 58786, 1352078, 208012, 5200300, 742900, 20058300, 2674440, 77558760, 9694845, 300540195, 35357670, 1166803110, 129644790, 4537567650
Offset: 0
Examples
This GCD equals A001405(n) for the smaller odd number gcd(C(12,6), C(11,5)) = gcd(924,462) = 462 = C(11,5).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Peter Luschny, Die schwingende Fakultät und Orbitalsysteme, August 2011.
- Peter Luschny, The lost Catalan numbers.
Programs
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Maple
A057977_ogf := proc(z) b := z -> (z-1)/(2*z^2); (2 + b(z))/sqrt(1-4*z^2) - b(z) end: seq(coeff(series(A057977_ogf(z),z,n+3),z,n), n = 0..35); A057977_rec := n -> `if`(n=0, 1, A057977_rec(n-1)*n^modp(n,2) *(4/(n+2))^modp(n+1,2)); A057977_int := proc(n) int((x^(2*n-1)*((4-x)^2/x)^cos(Pi*n))^(1/4),x=0..4)/(2*Pi); round(evalf(%)) end: A057977 := n -> (n!/iquo(n,2)!^2) / (iquo(n,2)+1): seq(A057977(n), n=0..35); # Peter Luschny, Apr 30 2011 b := proc(p, q) option remember; local S; if p = 0 and q = 0 then return 1 fi; if p < 0 or p > q then return 0 fi; S := b(p-2, q) + b(p, q-2); if type(q, odd) then S := S + b(p-1, q-1) fi; S end: seq(b(n, n), n=0..35); # Peter Luschny, Feb 23 2019
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Mathematica
a[n_] := n! / (Quotient[n, 2]!^2 * (Quotient[n, 2]+1)); Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Feb 03 2012, after Peter Luschny *)
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PARI
a(n)=if(n<0,0,(n+n%2)!/(n\2+1)!/(n\2+n%2)!/(1+n%2)) a(n)=n!/(n\2)!^2/(n\2+1) \\ Charles R Greathouse IV, May 02 2011
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Sage
def A057977(): x, n = 1, 1 while True: yield x m = n if is_odd(n) else 4/(n+2) x *= m n += 1 a = A057977(); [next(a) for i in range(36)] # Peter Luschny, Oct 21 2013
Formula
G.f.: (4*x^2+x-1+(1-x)*sqrt(1-4*x^2))/(2*sqrt(1-4*x^2)*x^2). E.g.f.: (1+1/x)*BesselI(1, 2*x). - Vladeta Jovovic, Jan 19 2004
From Peter Luschny, Apr 30 2011: (Start)
Recurrence: a(0) = 1 and a(n) = a(n-1)*n^[n odd]*(4/(n+2))^[n even] for n > 0.
Asymptotic formula: Let [n even] = 1 if n is even, 0 otherwise. Let N := n+1+[n even]. Then a(n) ~ 2^N /((n+1)^[n even]*sqrt(Pi*(2*N+1))).
Integral representation: a(n) = (1/(2*Pi))*Int_{x=0..4}(x^(2*n-1)* ((4-x)^2/x)^cos(Pi*n))^(1/4) (End)
E.g.f.: U(0) where U(k)= 1 + x/(1 - x/(x + (k+1)*(k+2)//U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 19 2012
From R. J. Mathar, Sep 16 2016: (Start)
D-finite with recurrence: (n+2)*a(n) - n*a(n-1) + 4*(-2*n+1)*a(n-2) + 4*(n-1)*a(n-3) + 16*(n-3)*a(n-4) = 0.
D-finite with recurrence: -(n+2)*(n^2-5)*a(n) + 4*(-2*n-1)*a(n-1) + 4*(n-1)*(n^2+2*n-4)*a(n-2) = 0. (End)
Sum_{n>=0} 1/a(n) = 8/3 + 8*Pi/(9*sqrt(3)). - Amiram Eldar, Aug 20 2022
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