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a(n) = n^floor(sqrt(n)).
1

%I #17 Sep 08 2022 08:45:24

%S 1,2,3,16,25,36,49,64,729,1000,1331,1728,2197,2744,3375,65536,83521,

%T 104976,130321,160000,194481,234256,279841,331776,9765625,11881376,

%U 14348907,17210368,20511149,24300000,28629151,33554432,39135393

%N a(n) = n^floor(sqrt(n)).

%C a(n) = n^(the integer part of the geometric mean of the divisors of n).

%C a(n) = n^[Card(k, 0<k<=n such that k is relatively prime to core(k)) where core(x) is the squarefree part of x].

%C a(n) = n^[Number of numbers k <=n with an odd number of divisors].

%H Robert Israel, <a href="/A117926/b117926.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = n^floor(sqrt(n)) = n^floor(n^(1/2)) = n^A000196(n).

%p seq(n^floor(sqrt(n)), n=1..100); # _Robert Israel_, Jun 22 2015

%t Table[n^Floor[Sqrt[n]], {n, 50}] (* _Vincenzo Librandi_, Jun 22 2015 *)

%o (Magma) [ n^ Floor(n^(1/2)) : n in [1..40]]; // _Zaki Khandaker_, Jun 21 2015

%o (PARI) vector(50, n, n^sqrtint(n)) \\ _Michel Marcus_, Jun 21 2015

%Y Cf. A000196.

%K easy,nonn

%O 1,2

%A _Jonathan Vos Post_, May 03 2006