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Minimal number of iterations of log_10 applied to n until the result is < 1.
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%I #18 May 16 2023 10:32:35

%S 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,

%T 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,

%U 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

%N Minimal number of iterations of log_10 applied to n until the result is < 1.

%C Different from A055642 and A138902, cf. Example.

%C Instead the real-valued log function one can consider only the integer part (i.e., A004216), since log_b(x) < k <=> x < b^k <=> floor(x) < b^k for any integer k >= 0; that's also why the first 2, 3, 4, ... appears exactly for 10, 10^10, 10^(10^10) etc. - _M. F. Hasler_, Dec 12 2018

%F With the definition 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:

%F a(E_{i=1..n} 10) = a(E_{i=1..n-1} 10) + 1, for n >= 1.

%F G.f.: g(x) = 1/(1-x)*Sum_{k>=0} x^(E_{i=1..k} 10).

%F = (x + x^10 + x^(10^10) + ...)/(1-x).

%e a(n) = 1, 2, 3, 4 for n = 1, 10, 10^10, 10^(10^10), i.e., n = 1, 10, 10000000000, 10^10000000000.

%e a(n) = 2 for all n >= 10, n < 10^10.

%t a[n_] := Length[NestWhileList[Log10, n, # >= 1 &]] - 1; Array[a, 100] (* _Amiram Eldar_, Dec 08 2018 *)

%o (PARI) a(n,i=1)={while(n=logint(n,10),i++);i} \\ _M. F. Hasler_, Dec 07 2018

%Y Cf. A001069, A010096, A211661, A211663, A211666, A211668, A211670.

%K base,nonn

%O 1,10

%A _Hieronymus Fischer_, Apr 30 2012

%E Name reworded by _M. F. Hasler_, Dec 12 2018