A384183 a(n) = minimum number of steps required to move n stones from a hole to the next one in an infinite row of holes, where at one step we can move any k stones at once from a hole to the hole at distance k to the left or to the right, and there are n stones overall.

0, 1, 2, 3, 4, 5, 5, 6, 7, 7, 6, 7, 7, 8, 7, 7, 8, 9, 8, 8, 9, 8, 8, 9, 8, 9, 8, 9, 9, 10, 9, 9, 10, 9, 9, 9, 10, 10, 9, 10, 9, 10, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 10, 10, 10, 10, 10, 11, 10, 11, 10, 11, 11, 10, 11, 11, 11, 11, 11, 11, 11, 11, 10, 11, 11, 11

Offset

0,3

Links

Fedor Karpelevitch, Table of n, a(n) for n = 0..113 (terms 0..43 from Fred Lunnon)

Fedor Karpelevitch, Sample solutions for n = 1..113 code here

Fedor Karpelevitch, All solutions (excluding permutations) for n=0..58

Fred Lunnon, A384183 sample solutions

Formula

a(n+k) <= a(n) + a(k) <= a(n) + k, for all n, k >= 0. - M. F. Hasler, May 24 2025

a(4k+1) <= 5+a(k+1), a(4k+2) <= 4+a(k+1), a(4k+3) <= 5+a(k+1), a(4k+4) <= 6+a(k+1) for all k > 0. - Konstantin Knop, May 27 2025

a(m^2*k+m) <= 2*m + a(k+1), for all k > 0 and m >= 2. - Yifan Xie, Jun 09 2025

Example

For n = 0 through n = 5, the fastest is to move n times 1 stone from the origin to the next hole.
For n = 6, one can move two times three stones three holes further, then three times two stones two holes "backwards", therefore a(6) = 2 + 3 = 5. Similarly, for a(3*4 = 12) = 3 + 4 = 7, and a(4*5 = 20) = 4 + 5 = 9. However, for larger oblong numbers, better solutions exist, and a(m*(m+1)) < 2m+1 for m > 4.

Crossrefs

Sequence in context: A138370 A125051 A064067 * A202306 A373178 A275579

Adjacent sequences: A384180 A384181 A384182 * A384184 A384185 A384186

Keyword

nonn,nice

Author

Konstantin Knop, May 21 2025

Extensions

a(11)-a(43) from Max Alekseyev, May 22 2025

Status

approved