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a(n) = 4*n*a(n-1) + 1 with a(0)=1.
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%I #41 Aug 15 2024 06:40:13

%S 1,5,41,493,7889,157781,3786745,106028861,3392923553,122145247909,

%T 4885809916361,214975636319885,10318830543354481,536579188254433013,

%U 30048434542248248729,1802906072534894923741,115385988642233275119425

%N a(n) = 4*n*a(n-1) + 1 with a(0)=1.

%C 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

%H Harvey P. Dale, <a href="/A056545/b056545.txt">Table of n, a(n) for n = 0..365</a>

%H Michael Z. Spivey and Laura L. Steil, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Spivey/spivey7.html">The k-Binomial Transforms and the Hankel Transform</a>, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.

%F a(n) = floor(e^(1/4)*4^n*n!).

%F From _Philippe Deléham_, Mar 14 2004: (Start)

%F a(n) = n!*Sum_{k=0..n} 4^(n-k)/k!.

%F E.g.f.: exp(x)/(1 - 4*x). (End)

%F a(n) = Sum_{k=0..n} P(n, k)*4^k. - _Ross La Haye_, Aug 29 2005

%F a(n) = hypergeometric_U(1, n+2 , 1/4)/4. - _Peter Luschny_, Nov 26 2014

%F 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

%F From _Peter Bala_, Mar 01 2017: (Start)

%F a(n) = Integral_{x = 0..inf} (4*x + 1)^n*exp(-x) dx.

%F The e.g.f. y = exp(x)/(1 - 4*x) satisfies the differential equation (1 - 4*x)*y' = (5 - 4*x)*y.

%F a(n) = (4*n + 1)*a(n-1) - 4*(n - 1)*a(n-2).

%F 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)

%e a(2) = 4*2*a(1) + 1 = 8*5 + 1 = 41.

%t 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 *)

%t nxt[{n_,a_}]:={n+1,4a(n+1)+1}; NestList[nxt,{0,1},20][[All,2]] (* _Harvey P. Dale_, Mar 19 2019 *)

%Y Cf. A000522, A010844, A010845, A056546, A056547, A001907 for analogs. A056545/(A000142*A000302) is an increasingly good approximation to 4th root of e.

%K nonn,easy

%O 0,2

%A _Henry Bottomley_, Jun 20 2000

%E More terms from _James A. Sellers_, Jul 04 2000