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%I A152577 #22 Apr 24 2021 09:21:21
%S A152577 11,1001,100001,10000001,1000000001,100000000001,10000000000001,
%T A152577 1000000000000001,100000000000000001,10000000000000000001,
%U A152577 1000000000000000000001,100000000000000000000001
%N A152577 a(n) = 10^(2*n - 1) + 1.
%C A152577 These numbers are all divisible by 11. This follows from the identity a^n - b^n = (a+b)*(a^(n-1) - a^(n-2)*b + ... + b^(n-1)) for odd values of n. In this example a=10 and b=1 so a+b = 11. The sum of digits rule for divisibility by 11 also applies.
%C A152577 Bisection of A000533. Also, bisection of A062397. a(n) is also A084508(n+1) written in base 2. a(n) is also A087289(n-1) written in base 2. a(n) is also the concatenation of "1", 2(n-1) digits "0" and "1". - _Omar E. Pol_, Dec 13 2008
%H A152577 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (101,-100).
%F A152577 a(n) = 100*a(n-1) - 99 (with a(1)=11). - _Vincenzo Librandi_, Dec 14 2010
%F A152577 G.f.: -11*x*(-1+10*x) / ( (100*x-1)*(x-1) ). - _R. J. Mathar_, Sep 01 2011
%F A152577 a(n) = 11*A095372(n-1). - _R. J. Mathar_, Sep 01 2011
%F A152577 a(n) = 101*a(n-1)-100*a(n-2). - _Wesley Ivan Hurt_, Apr 24 2021
%e A152577 From _Omar E. Pol_, Dec 14 2008: (Start)
%e A152577 n ....... a(n)
%e A152577 1 ....... 11
%e A152577 2 ...... 1001
%e A152577 3 ..... 100001
%e A152577 4 .... 10000001
%e A152577 5 ... 1000000001
%e A152577 (End)
%t A152577 LinearRecurrence[{101,-100},{11,1001},20] (* _Harvey P. Dale_, Nov 05 2015 *)
%o A152577 (PARI) g(n)=forstep(x=1,n,2,y=(10^x+1);print1(y","))
%Y A152577 Cf. A000533, A062397, A084508, A087289. - _Omar E. Pol_, Dec 13 2008
%K A152577 nonn,easy
%O A152577 1,1
%A A152577 _Cino Hilliard_, Dec 08 2008

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