cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

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%I A384183 #57 Jun 25 2025 00:26:52
%S A384183 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,
%T A384183 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,
%U A384183 10,11,10,11,10,11,11,10,11,11,11,11,11,11,11,11,10,11,11,11
%N 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.
%H A384183 Fedor Karpelevitch, <a href="/A384183/b384183.txt">Table of n, a(n) for n = 0..113</a> (terms 0..43 from Fred Lunnon)
%H A384183 Fedor Karpelevitch, <a href="/A384183/a384183_1.txt">Sample solutions for n = 1..113 </a> <a href="https://github.com/fedork/saigon_kt/tree/main">code here</a>
%H A384183 Fedor Karpelevitch, <a href="/A384183/a384183_2.txt">All solutions (excluding permutations) for n=0..58</a>
%H A384183 Fred Lunnon, <a href="/A384183/a384183.txt">A384183 sample solutions</a>
%F A384183 a(n+k) <= a(n) + a(k) <= a(n) + k, for all n, k >= 0. - _M. F. Hasler_, May 24 2025
%F A384183 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
%F A384183 a(m^2*k+m) <= 2*m + a(k+1), for all k > 0 and m >= 2. - _Yifan Xie_, Jun 09 2025
%e A384183 For n = 0 through n = 5, the fastest is to move n times 1 stone from the origin to the next hole.
%e A384183 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.
%K A384183 nonn,nice
%O A384183 0,3
%A A384183 _Konstantin Knop_, May 21 2025
%E A384183 a(11)-a(43) from _Max Alekseyev_, May 22 2025