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A259669 a(0)=0, a(1)=1, a(n) = min{5 a(k) + (5^(n-k)-1)/4, k=0..(n-1)} for n>=2.

%I A259669 #14 Jul 05 2015 03:10:59
%S A259669 0,1,6,11,36,61,86,211,336,461,586,1211,1836,2461,3086,3711,6836,9961,
%T A259669 13086,16211,19336,22461,38086,53711,69336,84961,100586,116211,131836,
%U A259669 209961,288086,366211,444336,522461,600586,678711,756836,1147461,1538086,1928711
%N A259669 a(0)=0, a(1)=1, a(n) = min{5 a(k) + (5^(n-k)-1)/4, k=0..(n-1)} for n>=2.
%C A259669 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=5.
%H A259669 Gheorghe Coserea, <a href="/A259669/b259669.txt">Table of n, a(n) for n = 0..4096</a>
%H A259669 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 A259669 P. Stockmeyer, <a href="http://www.cs.wm.edu/~pkstoc/boca.pdf">Variations on the Four-Post Tower of Hanoi Puzzle</a>
%F A259669 a(n) = min {5*a(k) + (5^(n-k)-1)/4 ; k < n}.
%F A259669 a(n) = sum(5^A003056(i), i=0..n-1).
%t A259669 a[n_] := a[n] = Min[ Table[ 5*a[k] + (5^(n-k) - 1)/4, {k, 0, n-1}]]; a[0] = 0; Table[a[n], {n, 0, 60}]
%Y A259669 Cf. A003056, A007664.
%K A259669 nonn
%O A259669 0,3
%A A259669 _Gheorghe Coserea_, Jul 02 2015