This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A152577 #26 Mar 14 2025 09:00:58 %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 %F A152577 E.g.f.: (exp(100*x) + 10*exp(x) - 11)/10. - _Stefano Spezia_, Mar 13 2025 %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