A259653 a(0)=0, a(1)=1, a(n) = min{3 a(k) + (3^(n-k)-1)/2, k=0..(n-1)} for n>=2.

0, 1, 4, 7, 16, 25, 34, 61, 88, 115, 142, 223, 304, 385, 466, 547, 790, 1033, 1276, 1519, 1762, 2005, 2734, 3463, 4192, 4921, 5650, 6379, 7108, 9295, 11482, 13669, 15856, 18043, 20230, 22417, 24604, 31165, 37726, 44287, 50848, 57409, 63970, 70531, 77092

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=3.

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.

Essentially partial sums of A098355.

Programs

Mathematica

a[n_] := a[n] = Min[ Table[ 3*a[k] + (3^(n-k) - 1)/2, {k, 0, n-1}]]; a[0] = 0; Table[a[n], {n, 0, 60}]

Formula

a(n) = min {3*a(k) + (3^(n-k)-1)/2 ; k < n}.

a(n) = sum(3^A003056(i), i=0..n-1).