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Gaussian binomial coefficients [ n,3 ] for q = 6.
2

%I #22 Sep 08 2022 08:44:46

%S 1,259,57535,12485095,2698853335,583026951031,125936508182839,

%T 27202382491194295,5875718100153221815,1269155234987097152695,

%U 274137535269957102205111,59213707780769522731688119,12790160886494733304250601655

%N Gaussian binomial coefficients [ n,3 ] for q = 6.

%H Vincenzo Librandi, <a href="/A022221/b022221.txt">Table of n, a(n) for n = 3..200</a>

%F G.f.: x^3/((1-x)*(1-6*x)*(1-36*x)*(1-216*x)). - _Vincenzo Librandi_, Aug 11 2016

%F a(n) = Product_{i=1..3} (6^(n-i+1)-1)/(6^i-1), by definition. - _Vincenzo Librandi_, Aug 11 2016

%t Table[QBinomial[n, 3, 6], {n, 3, 20}] (* _Vincenzo Librandi_, Aug 11 2016 *)

%o (Sage) [gaussian_binomial(n,3,6) for n in range(3,16)] # _Zerinvary Lajos_, May 27 2009

%o (Magma) r:=3; q:=6; [&*[(1-q^(n-i+1))/(1-q^i): i in [1..r]]: n in [r..20]]; // _Vincenzo Librandi_, Aug 11 2016

%o (PARI) r=3; q=6; for(n=r,30, print1(prod(j=1,r,(1-q^(n-j+1))/(1-q^j)), ", ")) \\ _G. C. Greubel_, Jun 07 2018

%o (GAP) List([3..15],n->Product([1..3],i->(6^(n-i+1)-1)/(6^i-1))); # _Muniru A Asiru_, Jul 04 2018

%K nonn,easy

%O 3,2

%A _N. J. A. Sloane_

%E Offset changed by _Vincenzo Librandi_, Aug 11 2016