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a(n) = a(n-1) + a(n-2) + n, a(0) = a(1) = 1.
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%I #28 Dec 29 2022 09:02:01

%S 1,1,4,8,16,29,51,87,146,242,398,651,1061,1725,2800,4540,7356,11913,

%T 19287,31219,50526,81766,132314,214103,346441,560569,907036,1467632,

%U 2374696,3842357,6217083,10059471,16276586,26336090,42612710,68948835,111561581

%N a(n) = a(n-1) + a(n-2) + n, a(0) = a(1) = 1.

%H Vincenzo Librandi, <a href="/A030119/b030119.txt">Table of n, a(n) for n = 0..400</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,-1,1).

%F Periodic mod 6.

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

%F a(n) = Lucas(n+2) + Fibonacci(n+1) - (n+3).

%F a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4); a(0)=1, a(1)=1, a(2)=4, a(3)=8. - _Harvey P. Dale_, Nov 06 2011

%F a(n) = -3 + (2^(-n)*((1-sqrt(5))^n*(-3+2*sqrt(5)) + (1+sqrt(5))^n*(3+2*sqrt(5)))) / sqrt(5) - n. - _Colin Barker_, Mar 11 2017

%t RecurrenceTable[{a[0]==a[1]==1,a[n]==a[n-1]+a[n-2]+n},a,{n,40}] (* or *) LinearRecurrence[{3,-2,-1,1},{1,1,4,8},40] (* _Harvey P. Dale_, Nov 06 2011 *)

%o (Magma) [Lucas(n+2) + Fibonacci(n+1) - (n+3) : n in [0..40]]; // _Vincenzo Librandi_, Nov 16 2011

%o (PARI) Vec((1-2*x+3*x^2-x^3)/((1-x-x^2)*(1-x)^2) + O(x^40)) \\ _Colin Barker_, Mar 11 2017

%o (PARI) vector(40, n, n--; f=fibonacci; f(n+3)+2*f(n+1)-n-3) \\ _G. C. Greubel_, Jul 24 2019

%o (Sage) f=fibonacci; [f(n+3)+2*f(n+1)-n-3 for n in (0..40)] # _G. C. Greubel_, Jul 24 2019

%o (GAP) F:=Fibonacci;; List([0..40], n-> F(n+3)+2*F(n+1)-n-3); # _G. C. Greubel_, Jul 24 2019

%Y Cf. A000032, A000045.

%K nonn,easy

%O 0,3

%A _Dragan Stevanovic_

%E Description corrected and sequence extended by _Erich Friedman_