login
a(n) = 4*a(n-1) - n - 1, for n > 0, a(0) = 1.
0

%I #62 Feb 01 2023 08:11:16

%S 1,2,5,16,59,230,913,3644,14567,58258,233021,932072,3728275,14913086,

%T 59652329,238609300,954437183,3817748714,15270994837,61083979328,

%U 244335917291,977343669142,3909374676545,15637498706156,62549994824599,250199979298370,1000799917193453,4003199668773784

%N a(n) = 4*a(n-1) - n - 1, for n > 0, a(0) = 1.

%C Last digit (using 0 to 9) is of period 10: repeat [1, 2, 5, 6, 9, 0, 3, 4, 7, 8].

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-9,4).

%F a(n) = (2^(2*n+1) + 3*n + 7)/9.

%F a(n) = 6*a(n-1) - 9*a(n-2) + 4*a(n-3), n >= 3.

%F a(n) = a(n-1) + A007583(n-1).

%F a(n) = 2*a(n-1) + A014825(n-1).

%F G.f.: (-2*x^2 + 4*x - 1)/((x - 1)^2*(4*x - 1)). - _Thomas Scheuerle_, Feb 03 2022

%F a(n) = -1 + 5*a(n-1) - 4*a(n-2), n >= 2.

%F a(n) = 1 + A160156(n-1), n >= 1.

%t LinearRecurrence[{6, -9, 4}, {1, 2, 5}, 28] (* _Amiram Eldar_, Feb 03 2022 *)

%o (PARI) a(n) = if (n, 4*a(n-1) - n - 1, 1); \\ _Michel Marcus_, Feb 03 2022

%o (Python)

%o print([(2**(2*n+1) + 3*n + 7)//9 for n in range(30)])

%o # _Gennady Eremin_, Feb 05 2022

%Y Cf. A007583 (first differences), A014825, A160156.

%K nonn,easy

%O 0,2

%A _Paul Curtz_, Feb 03 2022

%E More terms from _Michel Marcus_, Feb 03 2022