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a(n) = (11*n+5)*(n+4)*(n+3)*(n+2)*(n+1)/120.
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%I #20 Sep 08 2022 08:45:01

%S 1,16,81,266,686,1512,2982,5412,9207,14872,23023,34398,49868,70448,

%T 97308,131784,175389,229824,296989,378994,478170,597080,738530,905580,

%U 1101555,1330056,1594971,1900486,2251096,2651616,3107192,3623312

%N a(n) = (11*n+5)*(n+4)*(n+3)*(n+2)*(n+1)/120.

%H Harvey P. Dale, <a href="/A056118/b056118.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F a(n) = (11*n+5)*binomial(n+4,4)/5.

%F G.f.: (1+10*x)/(1-x)^6.

%F a(0)=1, a(1)=16, a(2)=81, a(3)=266, a(4)=686, a(5)=1512; for n>5, a(n) = 6*a(n-1) -15*a(n-2) +20*a(n-3) -15*a(n-4) +6*a(n-5) -a(n-6). - _Harvey P. Dale_, Oct 18 2013

%F From _G. C. Greubel_, Jan 17 2020: (Start)

%F a(n) = 11*binomial(n+5,5) - 8*binomial(n+4,4).

%F E.g.f.: (360 +2760*x +3720*x^2 +1560*x^3 +235*x^4 +11*x^5)*exp(x)/120. (End)

%p seq( (11*n+5)*binomial(n+4, 4)/5, n=0..40); # _G. C. Greubel_, Jan 17 2020

%t Table[((11n+5)Times@@(n+Range[4]))/120,{n,0,40}] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1}, {1,16,81,266,686,1512}, 40] (* _Harvey P. Dale_, Oct 18 2013 *)

%t Table[11*Binomial[n+5,5] -8*Binomial[n+4,4], {n,0,40}] (* _G. C. Greubel_, Jan 17 2020 *)

%o (PARI) vector(41, n, (11*n-6)*binomial(n+3,4)/5 ) \\ _G. C. Greubel_, Jan 17 2020

%o (Magma) [(11*n+5)*Binomial(n+4, 4)/5: n in [0..40]]; // _G. C. Greubel_, Jan 17 2020

%o (Sage) [(11*n+5)*binomial(n+4, 4)/5 for n in (0..40)] # _G. C. Greubel_, Jan 17 2020

%o (GAP) List([0..40], n-> (11*n+5)*Binomial(n+4, 4)/5 ); # _G. C. Greubel_, Jan 17 2020

%Y Cf. A055268.

%K nonn,easy

%O 0,2

%A _Barry E. Williams_, Jul 04 2000