A211665 Minimal number of iterations of log_10 applied to n until the result is < 1.
1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
Offset: 1
Examples
a(n) = 1, 2, 3, 4 for n = 1, 10, 10^10, 10^(10^10), i.e., n = 1, 10, 10000000000, 10^10000000000. a(n) = 2 for all n >= 10, n < 10^10.
Programs
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Mathematica
a[n_] := Length[NestWhileList[Log10, n, # >= 1 &]] - 1; Array[a, 100] (* Amiram Eldar, Dec 08 2018 *)
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PARI
a(n,i=1)={while(n=logint(n,10),i++);i} \\ M. F. Hasler, Dec 07 2018
Formula
With E_{i=1..n} c(i) := c(1)^(c(2)^(c(3)^(...(c(n-1)^(c(n)))...))); E_{i=1..0} := 1; example: E_{i=1..3} 10 = 10^(10^10) = 10^10000000000, we have:
a(E_{i=1..n} 10) = a(E_{i=1..n-1} 10) + 1, for n >= 1.
G.f.: g(x) = (1/(1-x))*Sum_{k>=0} x^(E_{i=1..k} 10) = (x + x^10 + x^(10^10) + ...)/(1-x).
Extensions
Name reworded by M. F. Hasler, Dec 12 2018
Comments