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%I #21 Jul 05 2015 03:13:34
%S 0,1,5,9,25,41,57,121,185,249,313,569,825,1081,1337,1593,2617,3641,
%T 4665,5689,6713,7737,11833,15929,20025,24121,28217,32313,36409,52793,
%U 69177,85561,101945,118329,134713,151097,167481,233017,298553,364089,429625,495161
%N a(0)=0, a(1)=1, a(n) = min{4 a(k) + (4^(n-k)-1)/3, k=0..(n-1)} for n>=2.
%C A generalization of Frame-Stewart recurrence is a(0)=0, a(1)=1, a(n)=min{q*a(k) + (q^(n-k)-1)/(q-1), k=0..(n-1)} where n>=2 and q>1. The sequence of first differences is q^A003056(n). For q=2 we have the sequence A007664. The current sequence is generated for q=4.
%H Gheorghe Coserea, <a href="/A259665/b259665.txt">Table of n, a(n) for n = 0..4096</a>
%H Jonathan Chappelon and Akihiro Matsuura, <a href="http://arxiv.org/abs/1009.0146">On generalized Frame-Stewart numbers</a>, arXiv:1009.0146 [math.NT], 2010.
%H P. Stockmeyer, <a href="http://www.cs.wm.edu/~pkstoc/boca.pdf">Variations on the Four-Post Tower of Hanoi Puzzle</a>
%F a(n) = min {4*a(k) + (4^(n-k)-1)/3 ; k < n}.
%F a(n) = sum(4^A003056(i), i=0..n-1).
%t a[n_] := a[n] = Min[ Table[ 4*a[k] + (4^(n-k) - 1)/3, {k, 0, n-1}]]; a[0] = 0; Table[a[n], {n, 0, 60}]
%Y Cf. A003056, A007664.
%K nonn
%O 0,3
%A _Gheorghe Coserea_, Jul 02 2015
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