A056547 a(n) = 6*n*a(n-1) + 1 with a(0)=1.

1, 7, 85, 1531, 36745, 1102351, 39684637, 1666754755, 80004228241, 4320228325015, 259213699500901, 17108104167059467, 1231783500028281625, 96079113002205966751, 8070645492185301207085, 726358094296677108637651

Offset: 0

Author

Henry Bottomley, Jun 20 2000

Keywords

• nonn
• easy

Examples

			a(2) = 6*2*a(1) + 1 = 12*7 + 1 = 85.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..346

Crossrefs

Cf. A000522, A010844, A010845, A056545, A056546 for analogs. A056547/(A000142*A000400) is an increasingly good approximation to 6th root of e.

Programs

Mathematica

nxt[{n_,a_}]:={n+1,6a(n+1)+1}; NestList[nxt,{0,1},20][[;;,2]] (* Harvey P. Dale, Jul 17 2024 *)

Formula

a(n) = floor(e^(1/6)*6^n*n!).

a(n) = n!*Sum_{k=0..n} 6^(n-k)/k!. E.g.f.: exp(x)/(1 - 6*x). - Philippe Deléham, Mar 14 2004

From Peter Bala, Mar 01 2017: (Start)

a(n) = Integral_{x = 0..inf} (6*x + 1)^n*exp(-x) dx.

The e.g.f. y = exp(x)/(1 - 6*x) satisfies the differential equation (1 - 6*x)*y' = (7 - 6*x)*y.

a(n) = (6*n + 1)*a(n-1) - 6*(n - 1)*a(n-2).

The sequence b(n) := 6^n*n! also satisfies the same recurrence with b(0) = 1, b(1) = 6. This leads to the continued fraction representation a(n) = 6^n*n!*( 1 + 1/(6 - 6/(13 - 12/(19 - ... - (6*n - 6)/(6*n + 1) )))) for n >= 2. Taking the limit gives the continued fraction representation exp(1/6) = 1 + 1/(6 - 6/(13 - 12/(19 - ... - (6*n - 6)/((6*n + 1) - ... )))). Cf. A010844. (End)

Extensions

More terms from James Sellers, Jul 04 2000