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