OFFSET
0,2
COMMENTS
For positive n, a(n) equals 4^n times the permanent of the n X n matrix with (5/4)'s along the main diagonal and 1's everywhere else. - John M. Campbell, Jul 10 2011
LINKS
Harvey P. Dale, Table of n, a(n) for n = 0..365
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
FORMULA
a(n) = floor(e^(1/4)*4^n*n!).
From Philippe Deléham, Mar 14 2004: (Start)
a(n) = n!*Sum_{k=0..n} 4^(n-k)/k!.
E.g.f.: exp(x)/(1 - 4*x). (End)
a(n) = Sum_{k=0..n} P(n, k)*4^k. - Ross La Haye, Aug 29 2005
a(n) = hypergeometric_U(1, n+2 , 1/4)/4. - Peter Luschny, Nov 26 2014
a(n) = exp(1/4)*4^n*Gamma(n+1, 1/4). a(n) ~ sqrt(2*Pi)*4^n*n^(n+1/2)*exp(1/4-n). - Vladimir Reshetnikov, Oct 14 2016
From Peter Bala, Mar 01 2017: (Start)
a(n) = Integral_{x = 0..inf} (4*x + 1)^n*exp(-x) dx.
The e.g.f. y = exp(x)/(1 - 4*x) satisfies the differential equation (1 - 4*x)*y' = (5 - 4*x)*y.
a(n) = (4*n + 1)*a(n-1) - 4*(n - 1)*a(n-2).
The sequence b(n) := 4^n*n! also satisfies the same recurrence with b(0) = 1, b(1) = 4. This leads to the continued fraction representation a(n) = 4^n*n!*( 1 + 1/(4 - 4/(9 - 8/(13 - ... - (4*n - 4)/(4*n + 1) )))) for n >= 2. Taking the limit gives the continued fraction representation exp(1/4) = 1 + 1/(4 - 4/(9 - 8/(13 - ... - (4*n - 4)/((4*n + 1) - ... )))). Cf. A010844. (End)
EXAMPLE
a(2) = 4*2*a(1) + 1 = 8*5 + 1 = 41.
MATHEMATICA
Round@Table[Exp[1/4] 4^n Gamma[n + 1, 1/4], {n, 0, 20}] (* Round is equivalent to FullSimplify here, but is much faster; Vladimir Reshetnikov, Oct 14 2016 *)
nxt[{n_, a_}]:={n+1, 4a(n+1)+1}; NestList[nxt, {0, 1}, 20][[All, 2]] (* Harvey P. Dale, Mar 19 2019 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Henry Bottomley, Jun 20 2000
EXTENSIONS
More terms from James A. Sellers, Jul 04 2000
STATUS
approved