login
a(n) = 12*n - a(n-1), with n>1, a(1)=1.
8

%I #43 Aug 04 2024 20:19:28

%S 1,23,13,35,25,47,37,59,49,71,61,83,73,95,85,107,97,119,109,131,121,

%T 143,133,155,145,167,157,179,169,191,181,203,193,215,205,227,217,239,

%U 229,251,241,263,253,275,265,287,277,299,289,311,301,323,313,335,325,347

%N a(n) = 12*n - a(n-1), with n>1, a(1)=1.

%H Vincenzo Librandi, <a href="/A166526/b166526.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 From _Harvey P. Dale_, Aug 10 2011: (Start)

%F a(n) = a(n-1) + a(n-2) - a(n-3), a(1)=1, a(2)=23, a(3)=13, for n>3.

%F G.f.: x*(1+22*x-11*x^2)/((x-1)^2*(1+x)). (End)

%F E.g.f.: 8*exp(-x) + 3*(1 + 2*x)*exp(x) - 11. - _G. C. Greubel_, May 16 2016

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 1/11 + (1/6 + 1/(4*sqrt(3)))*Pi. - _Amiram Eldar_, Feb 24 2023

%F a(n) = 6*n + 3 + 8*(-1)^n. - _G. C. Greubel_, Aug 04 2024

%t RecurrenceTable[{a[1]==1,a[n]==12n-a[n-1]},a[n],{n,60}] (* or *) LinearRecurrence[{1,1,-1},{1,23,13},60] (* _Harvey P. Dale_, Aug 10 2011 *)

%t CoefficientList[Series[(1 + 22 x - 11 x^2) / ((x - 1)^2 (1 + x)), {x, 0, 60}], x] (* _Vincenzo Librandi_, Sep 13 2013 *)

%o (Magma) [n eq 1 select 1 else 12*n-Self(n-1): n in [1..80]]; // _Vincenzo Librandi_, Sep 13 2013

%o (SageMath)

%o def A166526(n): return 6*n - 5 + 16*((n+1)%2)

%o [A166526(n) for n in range(1, 101)] # _G. C. Greubel_, Aug 04 2024

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

%K nonn,easy

%O 1,2

%A _Vincenzo Librandi_, Oct 16 2009