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%I #22 Mar 11 2024 23:11:19
%S 3,11,33,95,273,791,2313,6815,20193,60071,179193,535535,1602513,
%T 4799351,14381673,43112255,129271233,387682631,1162785753,3487832975,
%U 10462450353,31385253911,94151567433,282446313695,847322163873
%N a(n) is the only number m such that m = pi(1^(1/n)) + pi(2^(1/n)) + ... + pi(m^(1/n)).
%C We can easily prove that a(n) = 3^n + 2^n - 2.
%H Vincenzo Librandi, <a href="/A171270/b171270.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-11,6).
%F a(n) = 3^n + 2^n - 2.
%F G.f.: x*(3-7*x)/((1-3*x)*(1-2*x)*(1-x)). - _Vincenzo Librandi_, Mar 03 2014
%e pi(1) + pi(2) + pi(3)=3 so a(1)=3.
%p A171270:=n->3^n+2^n-2; seq(A171270(n), n=1..30); # _Wesley Ivan Hurt_, Feb 25 2014
%t Table[3^n+2^n-2,{n,26}]
%t CoefficientList[Series[(3 - 7 x)/((1 - 3 x) (1 - 2 x) (1 - x)), {x, 0, 30}], x] (* _Vincenzo Librandi_, Mar 03 2014 *)
%t LinearRecurrence[{6,-11,6},{3,11,33},30] (* _Harvey P. Dale_, Feb 28 2017 *)
%o (PARI) a(n)=3^n+2^n-2 \\ _Charles R Greathouse IV_, Jun 19 2013
%o (Magma) [3^n+2^n-2: n in [1..30]]; // _Vincenzo Librandi_, Mar 03 2014
%Y Cf. A000720.
%K easy,nice,nonn
%O 1,1
%A _Farideh Firoozbakht_, May 09 2010