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a(n) = n*10^n + 1.
6

%I #25 Sep 08 2022 08:45:04

%S 1,11,201,3001,40001,500001,6000001,70000001,800000001,9000000001,

%T 100000000001,1100000000001,12000000000001,130000000000001,

%U 1400000000000001,15000000000000001,160000000000000001,1700000000000000001,18000000000000000001

%N a(n) = n*10^n + 1.

%C Number of digits in (10^n)^(10^n) in base 10. - _Altug Alkan_, Apr 25 2016

%H Vincenzo Librandi, <a href="/A064748/b064748.txt">Table of n, a(n) for n = 0..1000</a>

%H Paul Leyland, <a href="http://www.leyland.vispa.com/numth/factorization/main.htm">Cullen and Woodall numbers and their generalization to other bases</a>

%F From _Ilya Gutkovskiy_, Apr 25 2016: (Start)

%F O.g.f.: (1 - 10*x + 90*x^2)/((1 - x)*(1 - 10*x)^2).

%F E.g.f.: (1 + 10*x*exp(9*x))*exp(x). (End)

%t Table[n 10^n + 1, {n, 0, 18}] (* _Michael De Vlieger_, Apr 25 2016 *)

%o (Magma) [ n*10^n+1: n in [0..20]]; // _Vincenzo Librandi_, Sep 16 2011

%o (PARI) a(n) = n*10^n + 1; \\ _Altug Alkan_, Apr 25 2016

%K easy,nonn

%O 0,2

%A _N. J. A. Sloane_, Oct 19 2001