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a(n) = 2*(n+1) + 3*n + ... + (k+1)*(n+2-k), where k = floor((n+1)/2).
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%I #19 Jul 12 2022 17:33:40

%S 4,6,17,22,43,52,86,100,150,170,239,266,357,392,508,552,696,750,925,

%T 990,1199,1276,1522,1612,1898,2002,2331,2450,2825,2960,3384,3536,4012,

%U 4182,4713,4902,5491,5700,6350,6580,7294,7546,8327,8602,9453,9752,10676,11000,12000

%N a(n) = 2*(n+1) + 3*n + ... + (k+1)*(n+2-k), where k = floor((n+1)/2).

%H G. C. Greubel, <a href="/A024305/b024305.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-3,-3,3,1,-1).

%F From _Vladeta Jovovic_, Jan 01 2003: (Start)

%F a(n) = (1/48)*(4*n^3 + (3*(-1)^(n+1) + 39)*n^2 + (18*(-1)^(n+1) + 74)*n + 27*(-1)^(n+1) + 27).

%F a(n) = a(n-1) + 3*a(n-2) - 3*a(n-3) - 3*a(n-4) + 3*a(n-5) + a(n-6) - a(n-7).

%F G.f.: x*(4 + 2*x - x^2 - x^3)/((1+x)^3*(1-x)^4). (End)

%F a(n) = Sum_{i=1..ceiling(n/2)} (i+1)*(n-i+2) = ceiling(n/2)*(-2*ceiling(n/2)^2 + 3n*ceiling(n/2) + 9*n + 14)/6. - _Wesley Ivan Hurt_, Sep 20 2013

%F E.g.f.: (1/24)*( x*(69 + 24*x + 2*x^2)*cosh(x) + (27 + 48*x + 27*x^2 + 2*x^3)*sinh(x) ). - _G. C. Greubel_, Jul 12 2022

%p seq(sum((i+1)*(k-i+2), i=1..ceil(k/2)), k=1..70); # _Wesley Ivan Hurt_, Sep 20 2013

%t Table[Ceiling[n/2]*(-2*Ceiling[n/2]^2+3n*Ceiling[n/2]+9n+14)/6,{n,100}] (* _Wesley Ivan Hurt_, Sep 20 2013 *)

%o (Magma)

%o b:= func< n | (1-(-1)^n)/2 >;

%o [(2*n^3 + 3*(6 +b(n))*n^2 + 2*(14 +9*b(n))*n + 27*b(n))/24 : n in [1..50]] // _G. C. Greubel_, Jul 12 2022

%o (SageMath)

%o def b(n): return (1-(-1)^n)/2

%o [(2*n^3 + 3*(6 +b(n))*n^2 + 2*(14 +9*b(n))*n + 27*b(n))/24 for n in (1..50)] # _G. C. Greubel_, Jul 12 2022

%Y Bisection: 2*A051925(n).

%Y Cf. A024854, A023855, A023856, A023857, A024868.

%K nonn

%O 1,1

%A _Clark Kimberling_

%E Name simplified by _Jon E. Schoenfield_, Jun 12 2019