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.

0, 1, 6, 11, 36, 61, 86, 211, 336, 461, 586, 1211, 1836, 2461, 3086, 3711, 6836, 9961, 13086, 16211, 19336, 22461, 38086, 53711, 69336, 84961, 100586, 116211, 131836, 209961, 288086, 366211, 444336, 522461, 600586, 678711, 756836, 1147461, 1538086, 1928711

Offset: 0

Author

Gheorghe Coserea, Jul 02 2015

Keywords

• nonn

Comments

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.

Links

• Gheorghe Coserea, Table of n, a(n) for n = 0..4096
• Jonathan Chappelon and Akihiro Matsuura, On generalized Frame-Stewart numbers, arXiv:1009.0146 [math.NT], 2010.
• P. Stockmeyer, Variations on the Four-Post Tower of Hanoi Puzzle

Crossrefs

Cf. A003056, A007664.

Programs

Mathematica

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}]

Formula

a(n) = min {5*a(k) + (5^(n-k)-1)/4 ; k < n}.

a(n) = sum(5^A003056(i), i=0..n-1).