A056546 a(n) = 5*n*a(n-1) + 1 with a(0)=1.

1, 6, 61, 916, 18321, 458026, 13740781, 480927336, 19237093441, 865669204846, 43283460242301, 2380590313326556, 142835418799593361, 9284302221973568466, 649901155538149792621, 48742586665361234446576

Offset

0,2

Links

Table of n, a(n) for n=0..15.

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^(1/5)*5^n*n!).

From Philippe Deléham, Mar 14 2004: (Start)

a(n) = n!*Sum_{k=0..n} 5^(n-k)/k!.

E.g.f.: exp(x)/(1 - 5*x). (End)

a(n) = Sum_{k=0..n} P(n, k)*5^k. - Ross La Haye, Aug 29 2005

a(n) = hypergeometric_U(1, n+2 , 1/5)/5. - Peter Luschny, Nov 26 2014

From Peter Bala, Mar 01 2017: (Start)

a(n) = Integral_{x >= 0} (5*x + 1)^n*exp(-x) dx.

The e.g.f. y = exp(x)/(1 - 5*x) satisfies the differential equation (1 - 5*x)*y' = (6 - 5*x)*y.

a(n) = (5*n + 1)*a(n-1) - 5*(n - 1)*a(n-2).

The sequence b(n) := 5^n*n! also satisfies the same recurrence with b(0) = 1, b(1) = 5. This leads to the continued fraction representation a(n) = 5^n*n!*( 1 + 1/(5 - 5/(11 - 10/(16 - ... - (5*n - 5)/(5*n + 1) )))) for n >= 2. Taking the limit gives the continued fraction representation exp(1/5) = 1 + 1/(5 - 5/(11 - 10/(16 - ... - (5*n - 5)/((5*n + 1) - ... )))). Cf. A010844. (End)

Example

a(2) = 5*2*a(1) + 1 = 10*6 + 1 = 61.

Mathematica

m = 16; CoefficientList[E^x/(1-5x) + O[x]^m, x] Range[0, m-1]! (* Jean-François Alcover, Jun 03 2019 *)

Crossrefs

Cf. A000522, A010844, A010845, A056545, A056547 for analogs. A056546/(A000142*A000351) is an increasingly good approximation to 5th root of e.

Sequence in context: A346983 A271841 A361526 * A127695 A144343 A022517

Adjacent sequences: A056543 A056544 A056545 * A056547 A056548 A056549

Keyword

nonn,easy

Author

Henry Bottomley, Jun 20 2000

Extensions

More terms from James A. Sellers, Jul 04 2000

Status

approved