OFFSET
1,3
COMMENTS
Starting with 2, n successive squarings yields an (a(n)+1)-digit number.
Dubickas proves that infinitely many terms of this sequence are divisible by 2 or 3 (and hence infinitely many composites). - Charles R Greathouse IV, Feb 04 2016
LINKS
Artūras Dubickas, Prime and composite integers close to powers of a number, Monatsh. Math. 158:3 (2009), pp. 271-284.
EXAMPLE
a(16) = 19728 because floor(log_10(2^(2^16))) = floor(log_10(2^65536)) = floor(log_10(2.003529930406846*10^19728)) = floor(19728.30179583467) = 19728.
MATHEMATICA
Table[Floor[Log[10, 2^(2^n)]], {n, 1, 29}] (* Vincenzo Librandi, Dec 30 2015 *)
PROG
(Magma) [Floor(Log(10, 2^(2^n))): n in [1..29]]; // Vincenzo Librandi, Dec 30 2015
(PARI) a(n) = floor(log(2^(2^n))/log(10)); \\ Michel Marcus, Dec 30 2015
(PARI) a(n)=logint(2^2^n, 10) \\ impractical except for small n, but avoids rounding; Charles R Greathouse IV, Feb 04 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, May 18 2007
STATUS
approved