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

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

Offset

0,2

Comments

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

References

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.

Links

Harry J. Smith, Table of n, a(n) for n = 0..200

Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Formula

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

Example

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)

Maple

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

Mathematica

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

Prog

(PARI) { for (n=0, 200, write("b059988.txt", n, " ", (10^n - 1)^2); ) } \\ Harry J. Smith, Jul 01 2009
(Python) def a(n): return (10**n - 1)**2  # Martin Gergov, Sep 10 2022

Crossrefs

Cf. A075411, A075412, A075413, A075414, A075415, A075416, A075417, A178630, A178631, A178632, A178633, A178634, A178635, A272066, A272067, A272068.

Subsequence of A238237.

Cf. A002275, A002283, A002477.

Sequence in context: A034993 A223328 A093281 * A017020 A237027 A185847

Adjacent sequences: A059985 A059986 A059987 * A059989 A059990 A059991

Keyword

easy,nonn

Author

Henry Bottomley, Mar 07 2001

Status

approved