A067314 Duplicate of A066022.
1, 1, 1, 2, 3, 4, 5, 6, 8, 9, 11, 12, 13, 15, 17, 18, 20, 21, 23, 25, 27, 28, 30, 32, 34, 35, 37
Offset: 0
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.
"Because some years previously in 1886 when occupied with the problem of the quadrature of the circle he had learned of the existence of a number computed to a relative degree of accuracy to be of such magnitude and of so many places, e.g., the 9th power of the 9th power of 9, that, the result having been obtained, 33 closely printed volumes of 1000 pages each of innumerable quires and reams of India paper would have to be requisitioned in order to contain the complete tale of its printed integers of units, tens, hundreds, thousands, tens of thousands, hundreds of thousands, millions, tens of millions, hundreds of millions, billions, the nucleus of the nebula of every digit of every series containing succinctly the potentiality of being raised to the utmost kinetic elaboration of any power of any of its powers." - James Joyce, Ulysses, Chapter 17. a(2)=2 since 2^(2^2)=2^4=16 has 2 digits. - _Carmine Suriano_, Feb 01 2011 a(0)=1 because 0^(0^0)=0^1=0, which has 1 digit. - _T. D. Noe_, Feb 01 2011
A055642 := proc(n) max(1,ilog10(n)+1) ; end proc: A054382 := proc(n) A055642(n^(n^n)) ; end proc: # R. J. Mathar, Feb 01 2011
f[ j_ ] := 1 + Floor[ Log[10, j] j^j ]; Table[ f[j], {j, 2, 20} ]
a(0) = 0, a(1) = 1, a(2) = 1 because 2^(2^2) = 16, a(3) = 7 because 3^(3^3) = 7625597484987 and its initial digit is 7, etc.
g[n_] := Quotient[n^p, 10^(Floor[ p*Log10@ n] - (1004 + p))]; f[n_] := Block[{p = n}, Quotient[ Nest[ g@ # &, p, p], 10^(1004 + p)]]; Array[f, 105, 0]
a:= n-> `if`(n=0, 1, (t-> floor(t/LambertW(t)))(n*log(10))): seq(a(n), n=0..80); # Alois P. Heinz, Nov 07 2019
a(n) = floor(solve(x=1, max(10,n+1), x^x-10^n))
a(n) = {my(k=1); while(k^k <= 10^n, k++); k-1} \\ Andrew Howroyd, Nov 07 2019
For k=1 to 9, k^k has k digits, that is, A066022(k) is a repdigit. k=631296394 is a term since k^k has 5555555555 digits. See Cobeli link.
Flatten@ Reap[Sow[0]; Do[v = d (10^nd-1)/9; s = Solve[v-1 <= x Log10[x] < v, x, Integers]; If[s != {}, Sow[x /. s]], {nd, 15}, {d, 9}]][[2, 1]] (* Giovanni Resta, Dec 05 2019 *)
isok(k) = #Set(digits(#Str(k^k))) == 1;
Flatten@ Reap[Sow[1]; Do[v = d (10^nd - 1)/9; s = Solve[v-1 <= x Log10[x] < v, x, Integers]; If[s != {}, Sow[v /. s]], {nd, 15}, {d, 9}]][[2, 1]] (* Giovanni Resta, Dec 05 2019 *)
a(10) = 1 + floor(10^(10+1) * log_10(10)) = 1 + floor( 100000000000 * 1) = 100000000001. a(10^3) = 3*10^3003 + 1.
Table[IntegerLength[(n^n)^(n^n)], {n, 1, 8}] (* Human friendly *)
Table[1 + Floor[n^(n + 1) * Log10[n]], {n, 1, 16}] (* Computationally efficient *)
a(n) = my(x=n^n); 1 + floor(x*log(x)/log(10));
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
For n=3, 3^3=11011_2 so a(3)=5.
[Floor(n*Log(n)/Log(2)) + 1: n in [1..70]]; // Vincenzo Librandi, Jul 15 2015
Array[IntegerLength[#^#, 2] &, 60] (* or *) Array[Floor[# Log[2, #]] + 1 &, 60] (* Michael De Vlieger, Jul 03 2015 *)
a(n) = #binary(n^n); \\ Michel Marcus, Jul 03 2015
Comments