A002282 a(n) = 8*(10^n - 1)/9.

0, 8, 88, 888, 8888, 88888, 888888, 8888888, 88888888, 888888888, 8888888888, 88888888888, 888888888888, 8888888888888, 88888888888888, 888888888888888, 8888888888888888, 88888888888888888, 888888888888888888, 8888888888888888888, 88888888888888888888, 888888888888888888888

Offset: 0

Author

N. J. A. Sloane

Keywords

• easy
• nonn

Comments

If the initial term is omitted, might be called eightful (or hateful) numbers!

Examples

			Curious multiplications:
9*9 + 7 = 88;
98*9 + 6 = 888;
987*9 + 5 = 8888;
9876*9 + 4 = 88888;
98765*9 + 3 = 888888;
987654*9 + 2 = 8888888;
9876543*9 + 1 = 88888888;
98765432*9 + 0 = 888888888;
987654321*9 - 1 = 8888888888;
9876543210*9 - 2 = 88888888888. - _Philippe Deléham_, Mar 09 2014
		

References

• Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 32.

Links

• Harry J. Smith, Table of n, a(n) for n = 0..200
• Index entries for linear recurrences with constant coefficients, signature (11,-10).

Crossrefs

Cf. A002275, A002276, A002277, A002278, A002279, A002280, A002281, A002283, A059988, A075412, A178635.

Cf. A010785, A017257, A051003, A059482, A246058.

Programs

Maple

A002282:=n->8*(10^n - 1)/9; seq(A002282(n), n=0..20); # Wesley Ivan Hurt, Mar 10 2014

Mathematica

LinearRecurrence[{11,-10}, {0,8}, 20] (* Harvey P. Dale, May 30 2013 *)

PARI

{ a=-4/5; for (n = 0, 200, a+=8*10^(n - 1); write("b002282.txt", n, " ", a); ) } \\ Harry J. Smith, Jun 27 2009

Python

def a(n): return 8*(10**n - 1)//9 # Martin Gergov, Oct 19 2022

Formula

From Jaume Oliver Lafont, Feb 03 2009: (Start)

a(n) = 11*a(n-1) - 10*a(n-2), with a(0)=0, a(1)=8.

G.f.: 8*x/((1-x)*(1-10*x)). (End)

a(n) = A178635(n)/A002283(n). - Reinhard Zumkeller, May 31 2010

a(n) = a(n-1) + 8*10^(n-1), with a(0)=0. - Vincenzo Librandi, Jul 22 2010

a(n) = 8*A002275(n) = A002283(n) - A002275(n). - Carauleanu Marc, Sep 03 2016

From Ilya Gutkovskiy, Sep 03 2016: (Start)

E.g.f.: 8*(exp(9*x) - 1)*exp(x)/9.

a(n) = floor(8*10^n/9). (End)

From Elmo R. Oliveira, Jul 20 2025: (Start)

a(n) = (A246058(n) - 1)/2.

a(n) = A010785(A017257(n-1)) for n >= 1. (End)