A246058 -id:A246058 - cp's OEIS frontend

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

N. J. A. Sloane

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

			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

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

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

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

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

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

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

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

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

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)

A246057 a(n) = (5*10^n - 2)/3.

Original entry on oeis.org

1, 16, 166, 1666, 16666, 166666, 1666666, 16666666, 166666666, 1666666666, 16666666666, 166666666666, 1666666666666, 16666666666666, 166666666666666, 1666666666666666, 16666666666666666, 166666666666666666, 1666666666666666666, 16666666666666666666, 166666666666666666666

Offset: 0

Vincenzo Librandi, Aug 13 2014

a(k-1) = (10^k - 4)/6, together with b(k) = 3*a(k-1) + 2 = A093143(k) and c(k) = 2*a(k-1) + 1 = A002277(k) are k-digit numbers for k >= 1 satisfying the so-called curious cubic identity a(k-1)^3 + b(k)^3 + c(k)^3 = a(k)*10^(2*k) + b(k)*10^k + c(k) (concatenated a(k)b(k)c(k)). This k-family and the proof of the identity has been given in the introduction of the van der Poorten reference. Thanks go to S. Heinemeyer for bringing these identities to my attention. - Wolfdieter Lang, Feb 07 2017

			Curious cubic identities (see a comment and reference above): 1^3 + 5^3 + 3^3 = 153, 16^3 + 50^3 + 33^3 = 165033, 166^3 + 500^3 + 333^3 = 166500333, ... - _Wolfdieter Lang_, Feb 07 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Alf van der Poorten, Kurt Thomsen, and Mark Wiebe, A curious cubic identity and self-similar sums of squares, The Mathematical Intelligencer, Vol. 29(2), pp. 39-41, March 2007.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. sequences with terms of the form 1k..k where the digit k is repeated n times: A000042 (k=1), A090843 (k=2), A097166 (k=3), A099914 (k=4), A099915 (k=5), this sequence (k=6), A246058 (k=7), A246059 (k=8), A067272 (k=9).

Cf. A002277, A086948, A093143, A323639.

Magma
```
[(5*10^n-2)/3: n in [0..20]];
```
Mathematica
```
Table[(5 10^n - 2)/3, {n, 0, 20}]
```

vector(50, n, (5*10^(n-1)-2)/3) \\ Derek Orr, Aug 13 2014

G.f.: (1 + 5*x)/((1 - x)*(1 - 10*x)).
a(n) = 11*a(n-1) - 10*a(n-2).
E.g.f.: exp(x)*(5*exp(9*x) - 2)/3. - Stefano Spezia, May 02 2025
a(n) = A323639(n+1)/2 = A086948(n+1)/12. - Elmo R. Oliveira, May 07 2025

cp's OEIS Frontend

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

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