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a(n) = ceiling((3-sqrt(3))*4^(n-3)) + 1 for n>=2, a(1)=1.
1

%I #31 Apr 01 2024 06:13:42

%S 1,2,3,7,22,83,326,1300,5195,20776,83098,332387,1329543,5318166,

%T 21272659,85090631,340362521,1361450080,5445800316,21783201259,

%U 87132805033,348531220128,1394124880509,5576499522030,22305998088117

%N a(n) = ceiling((3-sqrt(3))*4^(n-3)) + 1 for n>=2, a(1)=1.

%C From a new version of the camel problem. The original camel problem is discussed by de Bondt. A camel can carry one banana at a time on his back. It is on a diet and therefore can only have one banana at a time in its stomach. As soon as it has eaten a banana it walks a mile and then needs a new banana (in order to be able to continue its itinerary).

%C Let there be a stock of N bananas at the border of the desert. How far can the camel penetrate into the desert? (Of course it can form new stocks with transported bananas.)

%C The new version: Find a(n), the number of bananas needed for the camel to penetrate into the desert at least n miles.

%D Michiel de Bondt, The Camel-Banana Problem, Nieuw Archief voor de Wiskunde, 14-4, No. 3, 1996, pp. 415-426.

%D Matthijs Coster, Camels and Bananas, Preprint, Apr 29, 2004

%H Michiel de Bondt, <a href="https://arxiv.org/abs/2403.19667">The Camel-Banana Problem</a>, arXiv:2403.19667 [math.HO], 2024.

%H Matthijs Coster, <a href="http://www.coster.demon.nl/sequences/index.html">Sequences</a>

%t With[{s = Sqrt[3]}, MapAt[# - 1 &, Array[Ceiling[(3 - s)*4^(# - 3)] + 1 &, 25], 1]] (* _Michael De Vlieger_, Dec 07 2020 *)

%o (PARI) a(n) = if(n<=1,n==1,ceil((3-sqrt(3))*4^(n-3)) + 1); /* _Joerg Arndt_, Oct 20 2012 */

%K nonn,easy

%O 1,2

%A _Matthijs Coster_, Apr 29 2004

%E More terms from _Joshua Zucker_, May 03 2006