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%I #42 Feb 18 2024 08:22:28
%S 1,1,5,5,25,25,125,125,625,625,3125,3125,15625,15625,78125,78125,
%T 390625,390625,1953125,1953125,9765625,9765625,48828125,48828125,
%U 244140625,244140625,1220703125,1220703125,6103515625,6103515625
%N Inverse BinomialMean transform of the Fibonacci sequence A000045 (with the initial 0 omitted).
%C See A075271 for the definition of the BinomialMean transform.
%C The inverse binomial transform of 2^n*c(n+1), where c(n) is the solution to c(n) = c(n-1) + k*c(n-2), a(0)=0, a(1)=1 is 1, 1, 4k+1, 4k+1, (4k+1)^2, ... - _Paul Barry_, Feb 12 2004
%H Vincenzo Librandi, <a href="/A074872/b074872.txt">Table of n, a(n) for n = 1..2000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,5).
%F a(n) = 5^floor((n-1)/2).
%F a(1)=1, a(2)=1 and, for n > 2, a(n) = 5*a(n-2).
%F From _Paul Barry_, Feb 12 2004: (Start)
%F G.f.: x*(1+x)/(1-5*x^2);
%F a(n) = (1/(2*sqrt(5))*((1+sqrt(5))*(sqrt(5))^n - (1-sqrt(5))*(-sqrt(5))^n)).
%F Inverse binomial transform of A063727 (2^n*Fibonacci(n+1)). (End)
%F a(n+3) = a(n+2)*a(n+1)/a(n). - _Reinhard Zumkeller_, Mar 04 2011
%t a[1] := 1; a[2] := 1; a[n_] := 5a[n - 2]; Table[a[n], {n, 30}] (* _Alonso del Arte_, Mar 04 2011 *)
%o (Magma) [5^Floor((n-1)/2): n in [1..40]]; // _Vincenzo Librandi_, Aug 16 2011
%o (PARI) a(n)=5^((n-1)\2) \\ _Charles R Greathouse IV_, Oct 03 2016
%Y Cf. A000045, A075271, A056451, A016116, A108411.
%K nonn,easy
%O 1,3
%A _John W. Layman_, Sep 12 2002
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