A262695 a(n)=0 if n is in A259934, otherwise 1 + number of steps to reach the farthest leaf in that finite branch of the tree defined by edge-relation A049820(child) = parent.
0, 4, 0, 3, 2, 2, 0, 1, 1, 24, 3, 23, 0, 1, 2, 22, 2, 21, 0, 1, 1, 20, 0, 19, 1, 1, 3, 18, 1, 17, 0, 16, 2, 1, 0, 15, 1, 1, 10, 14, 1, 2, 0, 1, 2, 13, 0, 12, 9, 1, 1, 11, 1, 10, 0, 1, 1, 9, 0, 8, 8, 7, 0, 1, 1, 6, 1, 1, 1, 5, 0, 4, 7, 3, 1, 1, 13, 2, 0, 1, 2, 12, 4, 11, 6, 1, 3, 10, 1, 5, 0, 9, 2, 4, 0, 8, 5, 7, 1, 3, 1, 2, 0, 1, 4, 6, 0, 5, 1, 1, 2, 4, 1, 1, 0, 3, 1, 1, 0, 2, 3
Offset: 0
Keywords
Examples
For n=1, its transitive closure (as defined by edge-relation A049820(child) = parent) is the union of {1} itself together with all its descendants: {1, 3, 4, 5, 7, 8}. We see that there are no other nodes in this subtree whose root is 1, because A049820(3) = 3 - d(3) = 1, A049820(4) = 1, A049820(5) = 3, A049820(7) = 5, A049820(8) = 4 and of these only 7 and 8 are terms of A045765 (leaves). Starting iterating from 7 with A049820, we get 7 -> 5, 5 -> 3, 3 -> 1, and starting from 8 we get 8 -> 4, 4 -> 1, of which the former path is longer (3 steps), thus a(1) = 3+1 = 4.
For n=9, its transitive closure is {9, 11, 13, 15, 16, 17, 19, 21, 23, 24, 27, 29, 31, 33, 35, 36, 37, 39, 41, 43, 45, 47, 51, 53, 55, 57, 59, 61, 63, 64, 65, 67, 69, 71, 73, 75, 77, 79}. In this case the longest path is obtained by starting iterating from the largest of these: 79 -> 77 -> 73 -> 71 -> 69 -> 65 -> 61 -> 59 -> 57 -> 53 -> 51 -> 47 -> 45 -> 39 -> 35 -> 31 -> 29 -> 27 -> 23 -> 21 -> 17 -> 15 -> 11 -> 9, which is 23 steps long, thus a(9) = 23+1 = 24.
Links
- Antti Karttunen, Table of n, a(n) for n = 0..17724
Crossrefs
Formula
then a(n) = 0,
then a(n) = 1,
otherwise:
(In the last clause [ ] stands for Iverson bracket, giving as its result 1 only when A049820(k) = n, and 0 otherwise).