%I #44 Mar 08 2022 13:34:21
%S 0,0,0,0,0,0,720,5040,20160,60480,151200,332640,665280,1235520,
%T 2162160,3603600,5765760,8910720,13366080,19535040,27907200,39070080,
%U 53721360,72681840,96909120,127512000,165765600,213127200,271252800,342014400,427518000,530122320
%N Product of 6 consecutive integers.
%H Vincenzo Librandi, <a href="/A053625/b053625.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).
%F a(n) = n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5) = n!/(n-6)! = A052787(n)*(n-6) = a(n-1)*n/(n-6).
%F E.g.f.: x^6*exp(x).
%F a(n) = 720 * A000579(n). - _Zerinvary Lajos_, Apr 26 2007
%F For n > 5: a(n) = A173333(n, n-6). - _Reinhard Zumkeller_, Feb 19 2010
%F G.f.: 720*x^6/(1-x)^7. - _Colin Barker_, Mar 27 2012
%F a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7). - _Vincenzo Librandi_, Apr 28 2012
%F From _Amiram Eldar_, Mar 08 2022: (Start)
%F Sum_{n>=6} 1/a(n) = 1/600.
%F Sum_{n>=6} (-1)^n/a(n) = 4*log(2)/15 - 661/3600. (End)
%p seq(combinat[numbperm](n, 6), n=0..31); # _Zerinvary Lajos_, Apr 26 2007
%t CoefficientList[Series[720*x^6/(1-x)^7,{x,0,30}],x] (* _Vincenzo Librandi_, Apr 28 2012 *)
%t Times@@@Partition[Range[-5,30],6,1] (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,0,0,0,0,0,720},30] (* _Harvey P. Dale_, Nov 13 2015 *)
%t Pochhammer[Range[30]-6, 6] (* _G. C. Greubel_, Aug 27 2019 *)
%o (Magma) I:=[0,0,0,0,0,0,720]; [n le 7 select I[n] else 7*Self(n-1) -21*Self(n-2)+35*Self(n-3)-35*Self(n-4)+21*Self(n-5)-7*Self(n-6) +Self(n-7): n in [1..30]]; // _Vincenzo Librandi_, Apr 28 2012
%o (PARI) a(n)=factorback([n-5..n]) \\ _Charles R Greathouse IV_, Oct 07 2015
%o (Sage) [rising_factorial(n-5,6) for n in (0..30)] # _G. C. Greubel_, Aug 27 2019
%o (GAP) F:=Factorial;; Concatenation([0,0,0,0,0,0], List([6..30], n-> F(n)/F(n-5) )); # _G. C. Greubel_, Aug 27 2019
%Y Cf. A002378, A007531, A045619, A052762, A052787.
%Y Cf. A000579, A173333.
%K easy,nonn
%O 0,7
%A _Henry Bottomley_, Mar 20 2000