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a(n) = (10^n - 1)^2.
20

%I #75 Aug 01 2024 01:26:11

%S 0,81,9801,998001,99980001,9999800001,999998000001,99999980000001,

%T 9999999800000001,999999998000000001,99999999980000000001,

%U 9999999999800000000001,999999999998000000000001,99999999999980000000000001,9999999999999800000000000001,999999999999998000000000000001

%N a(n) = (10^n - 1)^2.

%C From _James D. Klein_, Feb 05 2012: (Start)

%C 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.:

%C 1/81 = 0.012345679...

%C 1/9801 = 0.000102030405060708091011...9799000102...

%C 1/998001 = 0.000001002003004005...997999000001002... (End)

%C Sum of first 10^n - 1 odd numbers. - _Arkadiusz Wesolowski_, Jun 12 2013

%D Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 32 at p. 61.

%D Walther Lietzmann, Lustiges und Merkwuerdiges von Zahlen und Formen, (F. Hirt, Breslau 1921-43), p. 149.

%H Harry J. Smith, <a href="/A059988/b059988.txt">Table of n, a(n) for n = 0..200</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (111,-1110,1000).

%F a(n) = 81*A002477(n) = A002283(n)^2 = (9*A002275(n))^2.

%F 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

%F a(n) = (A002283(n-1)*10 + 8) * 10^(n-1) + 1, for n>0. - _Reinhard Zumkeller_, May 31 2010

%F From _Ilya Gutkovskiy_, Apr 19 2016: (Start)

%F O.g.f.: 81*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).

%F E.g.f.: (1 - 2*exp(9*x) + exp(99*x))*exp(x). (End)

%F 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

%e From _Reinhard Zumkeller_, May 31 2010: (Start)

%e n=1: ..................... 81 = 9^2;

%e n=2: ................... 9801 = 99^2;

%e n=3: ................. 998001 = 999^2;

%e n=4: ............... 99980001 = 9999^2;

%e n=5: ............. 9999800001 = 99999^2;

%e n=6: ........... 999998000001 = 999999^2;

%e n=7: ......... 99999980000001 = 9999999^2;

%e n=8: ....... 9999999800000001 = 99999999^2;

%e n=9: ..... 999999998000000001 = 999999999^2. (End)

%p A059988:=n->(10^n-1)^2; seq(A059988(n), n=0..20); # _Wesley Ivan Hurt_, Nov 21 2013

%t Table[(10^n-1)^2, {n,0,20}] (* _Wesley Ivan Hurt_, Nov 21 2013 *)

%o (PARI) { for (n=0, 200, write("b059988.txt", n, " ", (10^n - 1)^2); ) } \\ _Harry J. Smith_, Jul 01 2009

%o (Python) def a(n): return (10**n - 1)**2 # _Martin Gergov_, Sep 10 2022

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

%Y Subsequence of A238237.

%Y Cf. A002275, A002283, A002477.

%K easy,nonn

%O 0,2

%A _Henry Bottomley_, Mar 07 2001