A033118 Base 8 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.

1, 8, 65, 520, 4161, 33288, 266305, 2130440, 17043521, 136348168, 1090785345, 8726282760, 69810262081, 558482096648, 4467856773185, 35742854185480, 285942833483841, 2287542667870728, 18300341342965825, 146402730743726600

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• base
• easy

Comments

Partial sums of A015565. - Mircea Merca, Dec 28 2010

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (8,1,-8).

Crossrefs

Pairwise sums are (8^n - 1)/7 (A023001).

Programs

Magma

[Round((8*8^n-8)/63): n in [1..30]]; // Vincenzo Librandi, Jun 25 2011

Maple

seq(1/7*floor(8^(n+1)/9),n=1..22); # Mircea Merca, Dec 27 2010

Mathematica

Table[FromDigits[PadRight[{},n,{1,0}],8],{n,20}] (* or *) LinearRecurrence[ {8,1,-8},{1,8,65},20] (* Harvey P. Dale, Jan 20 2021 *)

Formula

a(n) = 8*a(n-1) + a(n-2) - 8*a(n-3).

a(n) = 2^(3*n+3)/63 - 1/14 - (-1)^n/18. - R. J. Mathar, Jan 08 2011

From Paul Barry, Apr 04 2008: (Start)

G.f. x/((1-x^2)*(1-8*x));

a(n) = (1/3)*Sum_{k=0..n} A001045(3k). (End)

a(n) = floor(8^(n+1)/9)/7 = floor((8*8^n-1)/63) = round((8*8^n-8)/63) = round((16*8^n-9)/63) = ceiling((8*8^n-8)/63). a(n) = a(n-2) + 8^(n-1), n > 2. - Mircea Merca, Dec 28 2010