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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. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

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

FORMULA

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

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

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}]

CROSSREFS

Cf. A003056, A007664.

Sequence in context: A219500 A166702 A130667 * A108698 A002570 A038265

Adjacent sequences:  A259666 A259667 A259668 * A259670 A259671 A259672

KEYWORD

nonn

AUTHOR

Gheorghe Coserea, Jul 02 2015

STATUS

approved

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Last modified October 21 02:29 EDT 2018. Contains 316405 sequences. (Running on oeis4.)