A327603 Number of digits in (n^n)^(n^n).

1, 1, 3, 39, 617, 10922, 217833, 4871823, 121210687, 3327237897, 100000000001, 3268336354411, 115465060699617, 4386014250379643, 178300955775879752, 7725047653062230514, 355393490465494856466, 17303907095298306637188, 889028356166899850147118

Offset: 0

Author

Natan Arie Consigli, Sep 22 2019

Keywords

• nonn
• base

Comments

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

The standard simplification of (n^n)^(n^n) is n^(n^(n+1)). - M. F. Hasler, Oct 15 2019

Examples

			a(10) = 1 + floor(10^(10+1) * log_10(10)) = 1 + floor( 100000000000 * 1) = 100000000001.
a(10^3) = 3*10^3003 + 1.
		

Crossrefs

Cf. A004217, A054382, A055642, A066022.

Programs

Mathematica

Table[IntegerLength[(n^n)^(n^n)], {n, 1, 8}] (* Human friendly *)
Table[1 + Floor[n^(n + 1) * Log10[n]], {n, 1, 16}] (* Computationally efficient *)

PARI

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

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

Formula

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

a(10^k) = k * 10^(k*(10^k + 1)) + 1. - Jon E. Schoenfield, Sep 29 2019

a(n) = A055642(A004217(n)), n > 0. - Felix Fröhlich, Oct 15 2019

Extensions

a(9)-a(15) from Nathaniel Johnston, Sep 23 2019

a(13)-a(15) corrected and a(16) appended by Natan Arie Consigli, Sep 25 2019

a(17)-a(18) from Jon E. Schoenfield, Sep 29 2019