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a(n) = 5*a(n-1) - 2*n for n > 0, a(0) = 1.
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%I #20 Sep 12 2024 15:57:22

%S 1,3,11,49,237,1175,5863,29301,146489,732427,3662115,18310553,

%T 91552741,457763679,2288818367,11444091805,57220458993,286102294931,

%U 1430511474619,7152557373057,35762786865245,178813934326183,894069671630871,4470348358154309,22351741790771497,111758708953857435

%N a(n) = 5*a(n-1) - 2*n for n > 0, a(0) = 1.

%C In general, the ordinary generating function for the recurrence relation b(n) = k*b(n - 1) - m*n, with n > 0 and b(0)=1, is (1 - (m + 2)*x + x^2)/((1 - x)^2*(1 - k*x)). This recurrence gives the closed form b(n) = ((k^2 - k*(m + 2) + 1)*k^n + m*((k - 1)*n + k))/(k - 1)^2.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-11,5).

%F G.f.: (1 - 4*x + x^2)/((1 - x)^2*(1 - 5*x)).

%F a(n) = (4*n + 3*5^n + 5)/8.

%F Sum_{n>=0} 1/a(n) = 1.449934283402232875...

%F Lim_{n -> oo} a(n + 1)/a(n) = 5.

%F From _Elmo R. Oliveira_, Sep 10 2024: (Start)

%F E.g.f.: exp(x)*(3*exp(4*x) + 4*x + 5)/8.

%F a(n) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3) for n > 2. (End)

%t Table[(4 n + 3 5^n + 5)/8, {n, 0, 23}]

%t LinearRecurrence[{7, -11, 5}, {1, 3, 11}, 24]

%o (PARI) Vec((1-4*x+x^2)/((1-x)^2*(1-5*x)) + O(x^100)) \\ _Altug Alkan_, Feb 04 2016

%o (Magma) [(4*n + 3*5^n + 5)/8: n in [0..30]]; // _Vincenzo Librandi_, Feb 06 2016

%Y Cf. A014827, A024050, A094195, A104745, A107585, A164045, A176916, A221907.

%K nonn,easy

%O 0,2

%A _Ilya Gutkovskiy_, Feb 04 2016

%E a(24)-a(25) from _Elmo R. Oliveira_, Sep 10 2024