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Product_{i=1..3} (i+x) / Product_{i=1..3} (i-x) = Sum_{n>=0} (a(n)/b(n))*x^n.
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%I #48 Jan 01 2024 14:00:33

%S 1,11,121,971,6721,43331,269641,1648091,9981841,60176051,361921561,

%T 2174145611,13052763361,78340331171,470113403881,2820895001531,

%U 16926014399281,101558020876691,609353931324601,3656141011383851

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

%C If n mod 10 == 1, 2, or 4 then a(n)==0 (mod 11). - _Bruno Berselli_, Aug 26 2011

%H Vincenzo Librandi, <a href="/A068179/b068179.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (11,-36,36).

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

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

%F From _Yalcin Aktar_, Aug 10 2011: (Start)

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

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

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

%F 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)

%F G.f.: (1+36*x^2)/((1-2*x)*(1-3*x)*(1-6*x)). - _Bruno Berselli_, Aug 26 2011

%F E.g.f.: 10*exp(2*x) - 15*exp(3*x) + 6*exp(6*x). - _G. C. Greubel_, Nov 10 2018

%t Table[5*2^(n+1)+6^(n+1)-5*3^(n+1), {n,0,20}] (* _G. C. Greubel_, Nov 10 2018 *)

%t LinearRecurrence[{11,-36,36},{1,11,121},20] (* _Harvey P. Dale_, Aug 16 2021 *)

%o (Magma) [5*2^(n+1)+6^(n+1)-5*3^(n+1): n in [0..20]]; // _Vincenzo Librandi_, Aug 29 2011

%o (PARI) vector(20, n, n--; 5*2^(n+1)+6^(n+1)-5*3^(n+1)) \\ _G. C. Greubel_, Nov 10 2018

%o (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

%Y Cf. A026532.

%K nonn,frac,easy

%O 0,2

%A _Benoit Cloitre_, Mar 12 2002