A327603 - OEIS

%I #31 Jan 11 2020 15:57:47

%S 1,1,3,39,617,10922,217833,4871823,121210687,3327237897,100000000001,

%T 3268336354411,115465060699617,4386014250379643,178300955775879752,

%U 7725047653062230514,355393490465494856466,17303907095298306637188,889028356166899850147118

%N Number of digits in (n^n)^(n^n).

%C a(0) = 1 whether we take 0^0 = 1 or 0^0 = 0.

%C The standard simplification of (n^n)^(n^n) is n^(n^(n+1)). - _M. F. Hasler_, Oct 15 2019

%F a(n) = 1 + floor(n^(n+1) * log_10(n)).

%F a(10^k) = k * 10^(k*(10^k + 1)) + 1. - _Jon E. Schoenfield_, Sep 29 2019

%F a(n) = A055642(A004217(n)), n > 0. - _Felix Fröhlich_, Oct 15 2019

%e a(10) = 1 + floor(10^(10+1) * log_10(10)) = 1 + floor( 100000000000 * 1) = 100000000001.

%e a(10^3) = 3*10^3003 + 1.

%t Table[IntegerLength[(n^n)^(n^n)], {n, 1, 8}] (* Human friendly *)

%t Table[1 + Floor[n^(n + 1) * Log10[n]], {n, 1, 16}] (* Computationally efficient *)

%o (PARI) a(n) = my(x=n^n); 1 + floor(x*log(x)/log(10));

%o (PARI) A327603(n,L=log(10))=n^(n+1)*log(n)\L+1 \\ Supplying the 2nd arg allows to avoid re-computation of log(10) on each call, and also to get the number of digits in any desired base. - _M. F. Hasler_, Oct 15 2019

%Y Cf. A004217, A054382, A055642, A066022.

%K nonn,base

%O 0,3

%A _Natan Arie Consigli_, Sep 22 2019

%E a(9)-a(15) from _Nathaniel Johnston_, Sep 23 2019

%E a(13)-a(15) corrected and a(16) appended by _Natan Arie Consigli_, Sep 25 2019

%E a(17)-a(18) from _Jon E. Schoenfield_, Sep 29 2019