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a(n) = (1/2)*n*(5*n - 7); row 5 of A326728.
3

%I #13 Dec 24 2024 22:12:30

%S 0,-1,3,12,26,45,69,98,132,171,215,264,318,377,441,510,584,663,747,

%T 836,930,1029,1133,1242,1356,1475,1599,1728,1862,2001,2145,2294,2448,

%U 2607,2771,2940,3114,3293,3477,3666,3860,4059,4263,4472,4686,4905,5129,5358,5592

%N a(n) = (1/2)*n*(5*n - 7); row 5 of A326728.

%H Colin Barker, <a href="/A326725/b326725.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F From _Colin Barker_, Aug 04 2019: (Start)

%F G.f.: -x*(1 - 6*x)/(1 - x)^3.

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

%F E.g.f.: exp(x)*x*(5*x - 2)/2. - _Elmo R. Oliveira_, Dec 24 2024

%p a := n -> (1/2)*n*(5*n - 7): seq(a(n), n=0..48);

%o (PARI) concat(0, Vec(-x*(1 - 6*x) / (1 - x)^3 + O(x^40))) \\ _Colin Barker_, Aug 04 2019

%Y Cf. A326728.

%K sign,easy

%O 0,3

%A _Peter Luschny_, Aug 04 2019