A010845 - OEIS

A010845

a(n) = 3*n*a(n-1) + 1, a(0) = 1.

1, 4, 25, 226, 2713, 40696, 732529, 15383110, 369194641, 9968255308, 299047659241, 9868572754954, 355268619178345, 13855476147955456, 581929998214129153, 26186849919635811886, 1256968796142518970529

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

a(n)/(A000142*A000244) is an increasingly good approximation to cube root of e.

Related to Incomplete Gamma Function at 1/3. - Michael Somos, Mar 26 1999

For positive n, a(n) equals 3^n times the permanent of the n X n matrix with (4/3)'s along the main diagonal, and 1's everywhere else. - John M. Campbell, Jul 10 2011

REFERENCES

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 262.

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..200

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 262.

Roland Bacher, Counting Packings of Generic Subsets in Finite Groups, Electr. J. Combinatorics, 19 (2012), #P7. - From N. J. A. Sloane, Feb 06 2013

M. Z. Spivey and L. L. Steil, The k-Binomial Transforms and the Hankel Transform, J. Integ. Seqs. Vol. 9 (2006), #06.1.1.

FORMULA

E.g.f.: exp(x)/(1-3*x).

a(n) = floor( n!*e^(1/3)*3^n ) = n! * (Sum_{k=0..n} 3^(n-k) / k!) = n! * (e^(1/3) * 3^n - Sum_{k>n} 3^(n-k) / k!). - Michael Somos, Mar 26 1999

a(n) = Sum_{k=0..n} P(n, k)*3^k. - Ross La Haye, Aug 29 2005

Binomial transform of A032031. - Carl Najafi, Sep 11 2011

Conjecture: a(n) +(-3*n-1)*a(n-1) +3*(n-1)*a(n-2)=0. - R. J. Mathar, Feb 16 2014

a(n) = hypergeometric_U(1,n+2,1/3)/3. - Peter Luschny, Nov 26 2014

From Peter Bala, Mar 01 2017: (Start)

a(n) = Integral_{x >= 0} (3*x + 1)^n*exp(-x) dx.

The e.g.f. y = exp(x)/(1 - 3*x) satisfies the differential equation (1 - 3*x)*y' = (4 - 3*x)*y. Mathar's recurrence above follows easily from this.

The sequence b(n) := (3^n)*n! also satisfies Mathar's recurrence with b(0) = 1, b(1) = 3. This leads to the continued fraction representation a(n) = (3^n)*n!*( 1 + 1/(3 - 3/(7 - 6/(10 - ... - (3*n - 3)/(3*n + 1) )))) for n >= 2. Taking the limit as n -> oo gives the continued fraction representation exp(1/3) = 1 + 1/(3 - 3/(7 - 6/(10 - ... - (3*n - 3)/((3*n + 1) - ... )))). Cf. A010844. (End)

EXAMPLE

1 + 4*x + 25*x^2 + 226*x^3 + 2713*x^4 + 40696*x^5 + 732529*x^6 + ...

MATHEMATICA

Table[ Gamma[ n, 1/3 ]*Exp[ 1/3 ]*3^(n-1), {n, 1, 24} ]

a[ n_] := If[ n<0, 0, Floor[ n! E^(1/3) 3^n ]] (* Michael Somos, Sep 04 2013 *)

Range[0, 20]! CoefficientList[Series[Exp[x]/(1 - 3 x), {x, 0, 20}], x] (* Vincenzo Librandi, Feb 17 2014 *)

PROG

(PARI) {a(n) = if( n<0, 0, n! * sum(k=0, n, 3^(n-k) / k!))} /* Michael Somos, Sep 04 2013 */

CROSSREFS

Cf. A000522, A010844, A056545, A056546, A056547 for analogs.

Sequence in context: A340337 A001247 A031152 * A087660 A121660 A118835

Adjacent sequences: A010842 A010843 A010844 * A010846 A010847 A010848