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a(n) = (10*n + 7*(-1)^n + 5)/4.
1

%I #35 Aug 04 2024 20:19:38

%S 2,8,7,13,12,18,17,23,22,28,27,33,32,38,37,43,42,48,47,53,52,58,57,63,

%T 62,68,67,73,72,78,77,83,82,88,87,93,92,98,97,103,102,108,107,113,112,

%U 118,117,123,122,128,127,133,132,138,137,143,142,148,147,153,152,158

%N a(n) = (10*n + 7*(-1)^n + 5)/4.

%H Vincenzo Librandi, <a href="/A166539/b166539.txt">Table of n, a(n) for n = 1..1000</a>

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

%F a(n) = 5*n - a(n-1), n>=2.

%F From _Harvey P. Dale_, Jun 29 2011: (Start)

%F a(n) = a(n-1) + a(n-2) - a(n-3), n>=4.

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

%F From _G. C. Greubel_, May 16 2016: (Start)

%F E.g.f.: (1/4)*(5*(1 + 2*x)*exp(x) + 7*exp(-x) - 12).

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

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 1/3 + sqrt((5-2*sqrt(5))/5)*Pi/5. - _Amiram Eldar_, Feb 24 2023

%F a(n) = A166520(n) - (-1)^n. - _G. C. Greubel_, Aug 04 2024

%t RecurrenceTable[{a[1]==2,a[n]==5n-a[n-1]},a[n],{n,70}] (* or *) LinearRecurrence[{1,1,-1},{2,8,7},70] (* _Harvey P. Dale_, Jun 29 2011 *)

%o (Magma) [5*n/2 + (5+7*(-1)^n)/4: n in [1..70]]; // _Vincenzo Librandi_, May 15 2013

%o (SageMath)

%o def A166539(n): return (5*n - 1 + 7*((n+1)%2))//2

%o [A166539(n) for n in range(1, 101)] # _G. C. Greubel_, Aug 04 2024

%Y Cf. A166519, A166520, A166521, A166522, A166523, A166524, A166525, A166526.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Oct 16 2009