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%I #29 Apr 17 2024 09:46:37
%S 1,2,3,6,9,15,25,39,63,99,158,251,398,630,999,1584,2511,3981,6309,
%T 9999,15848,25118,39810,63095,99999,158489,251188,398107,630957,
%U 999999,1584893,2511886,3981071,6309573,9999999,15848931,25118864,39810717
%N Largest number whose 5th power has n digits.
%C Note that the rightmost digit of n and n^5 are identical. This is to 5th powers as A061439 is to cubes and A049416 is to squares.
%H Georg Fischer, <a href="/A114323/b114323.txt">Table of n, a(n) for n = 1..300</a> [First 152 terms by Vincenzo Librandi]
%F a(n) = ceiling((10^n)^(1/5)) - 1.
%e a(3) = 3 because 3^5 = 243 which has 3 digits, while 4^5 = 1024 has 3 digits.
%e a(32) = 2511886 because 2511886^5 = 99999914106500508412371346814176 has 32 digits, while 2511887^5 = 100000113160107495177704749808207 has 33 digits.
%p seq(print(n, floor((10^n-1)^(1/5))), n=1..300); # _Georg Fischer_ Apr 17 2024
%t Table[Floor[(10^n-1)^(1/5)],{n,40}] (* _Harvey P. Dale_, Dec 10 2012 *)
%o (PARI) a(n) = sqrtnint(10^n-1, 5) /* _Georg Fischer_ on proposal of _Michel Marcus_, Apr 16 2024 */
%o (Magma) [Ceiling((10^n)^(1/5))-1: n in [1..40]]; // _Vincenzo Librandi_, Oct 11 2011
%Y Cf. A061439, A049416.
%K easy,nonn,base,changed
%O 1,2
%A _Jonathan Vos Post_, Feb 06 2006
%E Data corrected by _Vincenzo Librandi_, Oct 11 2011
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