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a(n) = 17*n^2 - 1.
1

%I #21 Jul 03 2021 17:59:43

%S -1,16,67,152,271,424,611,832,1087,1376,1699,2056,2447,2872,3331,3824,

%T 4351,4912,5507,6136,6799,7496,8227,8992,9791,10624,11491,12392,13327,

%U 14296,15299,16336,17407,18512,19651,20824

%N a(n) = 17*n^2 - 1.

%C a(n) mod 9 = period 9: repeat [8, 7, 4, 8, 1, 1, 8, 4, 7] = A254375(n+5).

%C 1020 = 2*2*15*17.

%H Colin Barker, <a href="/A321180/b321180.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 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

%F a(n) = A244630(n) - 1.

%F a(n+1) = a(n) + 17*(2n+1).

%F a(n+10) = a(n) + 10*A061085(n+5).

%F a(n+15) = a(15-n) + 1020*n.

%F G.f.: (1 - 19*x - 16*x^2)/(-1 + x)^3. - _Stefano Spezia_, Oct 29 2018

%t a[n_]:=17*n^2 - 1; Array[a, 50] (* or *)

%t CoefficientList[Series[(1 - 19 x - 16 x^2)/(-1 + x)^3, {x, 0, 50}], x] (* _Stefano Spezia_, Oct 29 2018 *)

%t LinearRecurrence[{3,-3,1},{-1,16,67},40] (* _Harvey P. Dale_, Jul 03 2021 *)

%o (PARI) a(n)=17*n^2-1 \\ _Charles R Greathouse IV_, Oct 30 2018

%o (PARI) Vec((1 - 19*x - 16*x^2)/(-1 + x)^3 + O(x^50)) \\ _Colin Barker_, Oct 31 2018

%Y Cf. A000079, A005408, A061085, A244630, A254375.

%K sign,easy,less

%O 0,2

%A _Paul Curtz_, Oct 29 2018

%E One term corrected by _Colin Barker_, Oct 29 2018