OFFSET
1,2
FORMULA
Recurrence equation: a(n+1) = (2*n+2)*a(n) + a(n-1) with a(0) = 1 and a(1) = 1.
a(n) = Sum_{k = 0..floor((n-1)/2)} 2^(n-2*k-1)*(n-2*k-1)!*binomial(n-k-1,k)*binomial(n-k,k+1). Cf. A058798. - Peter Bala, Aug 01 2013
a(n) = 2^(n-1)*n!*hypergeometric([(1-n)/2, 1-n/2],[2, 1-n, -n], 1) for n>=2. - Peter Luschny, Sep 14 2014
MATHEMATICA
a[n_] := FromContinuedFraction[Range[0, 2n, 2]] // Numerator; Array[a, 20] (* Jean-François Alcover, Jun 03 2019 *)
PROG
(Sage)
def A036242(n):
if n == 1: return 1
return 2^(n-1)*factorial(n)*hypergeometric([1/2 - n/2, 1 - n/2], [2, 1-n, -n], 1)
[round(A036242(n).n(100)) for n in (1..20)] # Peter Luschny, Sep 14 2014
CROSSREFS
KEYWORD
frac,nonn,easy
AUTHOR
STATUS
approved