A350717 a(n) = 4*a(n-1) - n - 1, for n > 0, a(0) = 1.

1, 2, 5, 16, 59, 230, 913, 3644, 14567, 58258, 233021, 932072, 3728275, 14913086, 59652329, 238609300, 954437183, 3817748714, 15270994837, 61083979328, 244335917291, 977343669142, 3909374676545, 15637498706156, 62549994824599, 250199979298370, 1000799917193453, 4003199668773784

Offset: 0

Author

Paul Curtz, Feb 03 2022

Keywords

• nonn
• easy

Comments

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

Links

• Index entries for linear recurrences with constant coefficients, signature (6,-9,4).

Crossrefs

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

Programs

Mathematica

LinearRecurrence[{6, -9, 4}, {1, 2, 5}, 28] (* Amiram Eldar, Feb 03 2022 *)

PARI

a(n) = if (n, 4*a(n-1) - n - 1, 1); \\ Michel Marcus, Feb 03 2022

Python

print([(2**(2*n+1) + 3*n + 7)//9 for n in range(30)])
# Gennady Eremin, Feb 05 2022

Formula

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

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

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

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

G.f.: (-2*x^2 + 4*x - 1)/((x - 1)^2*(4*x - 1)). - Thomas Scheuerle, Feb 03 2022

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

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

Extensions

More terms from Michel Marcus, Feb 03 2022