A259665 - OEIS

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A259665

a(0)=0, a(1)=1, a(n) = min{4 a(k) + (4^(n-k)-1)/3, k=0..(n-1)} for n>=2.

0, 1, 5, 9, 25, 41, 57, 121, 185, 249, 313, 569, 825, 1081, 1337, 1593, 2617, 3641, 4665, 5689, 6713, 7737, 11833, 15929, 20025, 24121, 28217, 32313, 36409, 52793, 69177, 85561, 101945, 118329, 134713, 151097, 167481, 233017, 298553, 364089, 429625, 495161

(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=4.

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 {4*a(k) + (4^(n-k)-1)/3 ; k < n}.

a(n) = sum(4^A003056(i), i=0..n-1).

MATHEMATICA

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}]

CROSSREFS

Cf. A003056, A007664.

Sequence in context: A024825 A147074 A147192 * A284285 A025624 A147496