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a(n) = (5*n + 4)*binomial(n+7,7)/4.
2

%I #13 Sep 08 2022 08:45:01

%S 1,18,126,570,1980,5742,14586,33462,70785,140140,262548,469404,806208,

%T 1337220,2151180,3368244,5148297,7700814,11296450,16280550,23088780,

%U 32265090,44482230,60565050,81516825,108548856,143113608,186941656

%N a(n) = (5*n + 4)*binomial(n+7,7)/4.

%H G. C. Greubel, <a href="/A056125/b056125.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (9,-36,84,-126,126,-84,36,-9,1).

%F G.f.: (1+9*x)/(1-x)^9.

%F a(0)=1, a(1)=18, a(2)=126, a(3)=570, a(4)=1980, a(5)=5742, a(6)=14586, a(7)=33462, a(8)=70785, a(n) = 9*a(n-1) -36*a(n-2) +84*a(n-3) -126*a(n-4) + 126*a(n-5) -84*a(n-6) +36*a(n-7) -9*a(n-8) +a(n-9). - _Harvey P. Dale_, Jan 18 2013

%F From _G. C. Greubel_, Jan 19 2020: (Start)

%F a(n) = 10*binomial(n+8,8) - 9*binomial(n+7,7).

%F E.g.f.: (20160 + 342720*x + 917280*x^2 + 823200*x^3 + 323400*x^4 + 62328*x^5 + 6076*x^6 + 284*x^7 + 5*x^8)*exp(x)/20160. (End)

%p seq( (5*n+4)*binomial(n+7,7)/4, n=0..30); # _G. C. Greubel_, Jan 19 2020

%t Table[((5n+4)Binomial[n+7,7])/4,{n,0,30}] (* or *) LinearRecurrence[{9,-36,84, -126,126,-84,36,-9,1},{1,18,126,570,1980,5742,14586,33462,70785},30] (* _Harvey P. Dale_, Jan 18 2013 *)

%o (PARI) vector(31, n, (5*n-1)*binomial(n+6,7)/4 ) \\ _G. C. Greubel_, Jan 19 2020

%o (Magma) [(5*n+4)*Binomial(n+7,7)/4: n in [0..30]]; // _G. C. Greubel_, Jan 19 2020

%o (Sage) [(5*n+4)*binomial(n+7,7)/4 for n in (0..30)] # _G. C. Greubel_, Jan 19 2020

%o (GAP) List([0..30], n-> (5*n+4)*Binomial(n+7,7)/4 ); # _G. C. Greubel_, Jan 19 2020

%Y Cf. A052254.

%Y Cf. A093645 ((10, 1) Pascal, column m=8).

%Y Partial sums of A052254.

%K easy,nonn

%O 0,2

%A _Barry E. Williams_, Jul 07 2000