A068179 Product_{i=1..3} (i+x) / Product_{i=1..3} (i-x) = Sum_{n>=0} (a(n)/b(n))*x^n.

1, 11, 121, 971, 6721, 43331, 269641, 1648091, 9981841, 60176051, 361921561, 2174145611, 13052763361, 78340331171, 470113403881, 2820895001531, 16926014399281, 101558020876691, 609353931324601, 3656141011383851

Offset: 0

Author

Benoit Cloitre, Mar 12 2002

Keywords

• nonn
• frac
• easy

Comments

If n mod 10 == 1, 2, or 4 then a(n)==0 (mod 11). - Bruno Berselli, Aug 26 2011

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (11,-36,36).

Crossrefs

Cf. A026532.

Programs

Magma

[5*2^(n+1)+6^(n+1)-5*3^(n+1): n in [0..20]]; // Vincenzo Librandi, Aug 29 2011

Mathematica

Table[5*2^(n+1)+6^(n+1)-5*3^(n+1), {n,0,20}] (* G. C. Greubel, Nov 10 2018 *)
LinearRecurrence[{11,-36,36},{1,11,121},20] (* Harvey P. Dale, Aug 16 2021 *)

PARI

vector(20, n, n--; 5*2^(n+1)+6^(n+1)-5*3^(n+1)) \\ G. C. Greubel, Nov 10 2018

Python

for n in range(0,20): print(5*2**(n+1)+6**(n+1)-5*3**(n+1), end=', ') # Stefano Spezia, Nov 12 2018

Formula

b(n) = A026532(2*n-1) for n >= 1.

Lim_{n -> infinity} a(n)/b(n) = 12.

From Yalcin Aktar, Aug 10 2011: (Start)

a(n) = 5*2^(n+1) + 6^(n+1) - 5*3^(n+1).

a(n)/b(n) = 12 - 30/2^n + 20/3^n.

General case: lim_{m-->+oo} a_n(m)/b_n(m) = A002378(n) where

Product_{i=1..d} (x+i)/Product_{i=1..d} (i-x) = Sum_{n>=0} (a_d(n)/b_d(n))*x^n) = ((-1)^d) * (1 + Sum_{j>=1} (Sum_{k=1..d} ((-1)^k/k^j) * binomial(2*k,k) * binomial(d+k,2*k)) * x^j). (End)

G.f.: (1+36*x^2)/((1-2*x)*(1-3*x)*(1-6*x)). - Bruno Berselli, Aug 26 2011

E.g.f.: 10*exp(2*x) - 15*exp(3*x) + 6*exp(6*x). - G. C. Greubel, Nov 10 2018