A272068 -id:A272068 - cp's OEIS frontend

A059988 a(n) = (10^n - 1)^2.

0, 81, 9801, 998001, 99980001, 9999800001, 999998000001, 99999980000001, 9999999800000001, 999999998000000001, 99999999980000000001, 9999999999800000000001, 999999999998000000000001, 99999999999980000000000001, 9999999999999800000000000001, 999999999999998000000000000001

Offset: 0

Henry Bottomley, Mar 07 2001

From James D. Klein, Feb 05 2012: (Start)
The periods of the reciprocals of a(n) are the consecutive integers from 0 to 10^n-1, omitting the one integer 10^n-2, right-justified in field widths of size n.
E.g.:
  1/81 = 0.012345679...
  1/9801 = 0.000102030405060708091011...9799000102...
  1/998001 = 0.000001002003004005...997999000001002... (End)
Sum of first 10^n - 1 odd numbers. - Arkadiusz Wesolowski, Jun 12 2013

			From _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ..................... 81 = 9^2;
n=2: ................... 9801 = 99^2;
n=3: ................. 998001 = 999^2;
n=4: ............... 99980001 = 9999^2;
n=5: ............. 9999800001 = 99999^2;
n=6: ........... 999998000001 = 999999^2;
n=7: ......... 99999980000001 = 9999999^2;
n=8: ....... 9999999800000001 = 99999999^2;
n=9: ..... 999999998000000001 = 999999999^2. (End)

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 32 at p. 61.
Walther Lietzmann, Lustiges und Merkwuerdiges von Zahlen und Formen, (F. Hirt, Breslau 1921-43), p. 149.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 34.

Harry J. Smith, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A075411, A075412, A075413, A075414, A075415, A075416, A075417, A178630, A178631, A178632, A178633, A178634, A178635, A272066, A272067, A272068.
Subsequence of A238237.
Cf. A002275, A002283, A002477.

A059988:=n->(10^n-1)^2; seq(A059988(n), n=0..20); # Wesley Ivan Hurt, Nov 21 2013

Table[(10^n-1)^2, {n,0,20}] (* Wesley Ivan Hurt, Nov 21 2013 *)

a(n) = { (10^n - 1)^2 } \\ Harry J. Smith, Jul 01 2009

def a(n): return (10**n - 1)**2  # Martin Gergov, Sep 10 2022

a(n) = 81*A002477(n) = A002283(n)^2 = (9*A002275(n))^2.
a(n) = {999... (n times)}^2 = {999... (n times), 000... (n times)} - {999... (n times)}. For example, 999^2 = 999000 - 999 = 998001. - Kyle D. Balliet, Mar 07 2009
a(n) = (A002283(n-1)*10 + 8) * 10^(n-1) + 1, for n>0. - Reinhard Zumkeller, May 31 2010
From Ilya Gutkovskiy, Apr 19 2016: (Start)
O.g.f.: 81*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).
E.g.f.: (1 - 2*exp(9*x) + exp(99*x))*exp(x). (End)
Sum_{n>=1} 1/a(n) = (log(10)*(QPolyGamma(0, 1, 1/10) - log(10/9)) + QPolyGamma(1, 1, 1/10))/log(10)^2 = 0.012448721523422795191... . - Stefano Spezia, Jul 31 2024
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3). - Elmo R. Oliveira, Aug 02 2025

A272066 a(n) = (10^n-1)^3.

Original entry on oeis.org

0, 729, 970299, 997002999, 999700029999, 999970000299999, 999997000002999999, 999999700000029999999, 999999970000000299999999, 999999997000000002999999999, 999999999700000000029999999999, 999999999970000000000299999999999, 999999999997000000000002999999999999

Offset: 0

Seiichi Manyama, Apr 19 2016

The sum of the digits of a(n) is divisible by 18. For example, 9^3 = 729 and 7 + 2 + 9 = 18 * 1.

Number of 9 in a(n) is 2*n-1 for n > 0. - Seiichi Manyama, Sep 18 2018

			From _Seiichi Manyama_, Sep 18 2018: (Start)
n| a(n) can be divided into 3 parts for n > 1.
-+--------------------------------------------
1|        72    9
2|   9   702   99
3|  99  7002  999
4| 999 70002 9999
(End)

Index entries for linear recurrences with constant coefficients, signature (1111,-112110,1111000,-1000000).

Cf. A002283, A059988, A272067, A272068, A319358.

[(10^n-1)^3 : n in [0..10]]; // Wesley Ivan Hurt, Apr 19 2016

A272066:=n->(10^n-1)^3: seq(A272066(n), n=0..15); # Wesley Ivan Hurt, Apr 19 2016

(10^Range[0, 10] - 1)^3 (* Wesley Ivan Hurt, Apr 19 2016 *)

a(n) = (10^n-1)^3; \\ Michel Marcus, Apr 19 2016

Ruby
```
(0..n).each{|i| p ('9' * i).to_i ** 3}
```

a(n) = A002283(n)^3.
From Ilya Gutkovskiy, Apr 19 2016: (Start)
O.g.f.: 729*x*(1 + 220*x + 1000*x^2)/((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)).
E.g.f.: (-1 + 3*exp(9*x) - 3*exp(99*x) + exp(999*x))*exp(x). (End)

A272067 a(n) = (10^n-1)^4.

Original entry on oeis.org

0, 6561, 96059601, 996005996001, 9996000599960001, 99996000059999600001, 999996000005999996000001, 9999996000000599999960000001, 99999996000000059999999600000001, 999999996000000005999999996000000001, 9999999996000000000599999999960000000001, 99999999996000000000059999999999600000000001

Offset: 0

Seiichi Manyama, Apr 19 2016

The sum of the digits of a(n) is divisible by 18. For example, 9^4 = 6561 and 6 + 5 + 6 + 1 = 18 * 1.

Number of 9 in a(n) is 2*n-2 for n > 0. - Seiichi Manyama, Sep 18 2018

			From _Seiichi Manyama_, Sep 18 2018: (Start)
n| a(n) can be divided into 4 parts for n > 1.
-+--------------------------------------------
1|        65        61
2|   9   605   9   601
3|  99  6005  99  6001
4| 999 60005 999 60001
(End)

Index entries for linear recurrences with constant coefficients, signature (11111,-11222110,1122211000,-11111000000,10000000000).

Cf. A002283, A059988, A272066, A272068, A319358.

[(10^n-1)^4 : n in [0..10]]; // Wesley Ivan Hurt, Apr 19 2016

A272067:=n->(10^n-1)^4: seq(A272067(n), n=0..15); # Wesley Ivan Hurt, Apr 19 2016

(10^Range[0, 10] - 1)^4 (* Wesley Ivan Hurt, Apr 19 2016 *)

a(n) = (10^n-1)^4; \\ Michel Marcus, Apr 19 2016

Ruby
```
(0..n).each{|i| p ('9' * i).to_i ** 4}
```

a(n) = A059988(n)^2 = A002283(n)^4.
From Ilya Gutkovskiy, Apr 19 2016: (Start)
O.g.f.: 6561*x*(1 + 100*x)*(1 + 3430*x + 10000*x^2)/((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)*(1 - 10000*x)).
E.g.f.: (1 - 4*exp(9*x) + 6*exp(99*x) - 4*exp(999*x) + exp(9999*x))*exp(x). (End)

A319358 a(n) = (10^n - 1)^9.

Original entry on oeis.org

0, 387420489, 913517247483640899, 991035916125874083964008999, 999100359916012598740083996400089999, 999910003599916001259987400083999640000899999, 999991000035999916000125999874000083999964000008999999

Offset: 0

Seiichi Manyama, Sep 17 2018

nonn

Number of 9 in a(n) is 5*n-8 for n > 2.

			n|   a(n) can be divided into 9 parts for n > 3.
-+------------------------------------------------------
3|   99   1035   9   16125      874083   9   64008   999
4|  999  10035  99  160125  9  8740083  99  640008  9999
5| 9999 100035 999 1600125 99 87400083 999 6400008 99999

Seiichi Manyama, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (1111111111, -112233445443322110, 1123457901110987543211000, -1123570145779775409653211000000, 112358025801220975197532110000000000, -1123570145779775409653211000000000000000, 1123457901110987543211000000000000000000000, -112233445443322110000000000000000000000000000, 1111111111000000000000000000000000000000000000, -1000000000000000000000000000000000000000000000).

Cf. A001017, A002283, A059988, A272067, A272068.

a[n_]:=(10^n - 1)^9 ; Array[a,  50, 0] (* Stefano Spezia, Sep 17 2018 *)

PARI
```
a(n) = (10^n-1)^9;
```

a(n) = A002283(n)^9.

cp's OEIS Frontend

A059988 a(n) = (10^n - 1)^2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

A272066 a(n) = (10^n-1)^3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Ruby

Formula

A272067 a(n) = (10^n-1)^4.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Ruby

Formula

A319358 a(n) = (10^n - 1)^9.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula