A213725 -id:A213725 - cp's OEIS frontend

A230425 a(n)=0 if n is in the infinite trunk of factorial beanstalk (in A219666), otherwise 1 + number of steps to reach the farthest leaf in that finite branch of the beanstalk.

0, 0, 0, 1, 1, 0, 2, 0, 1, 1, 0, 3, 0, 1, 1, 2, 2, 0, 1, 1, 1, 1, 1, 0, 2, 0, 1, 1, 0, 3, 0, 1, 1, 2, 3, 0, 1, 1, 2, 2, 0, 1, 1, 1, 1, 1, 0, 3, 0, 1, 1, 2, 0, 3, 1, 1, 3, 0, 2, 1, 1, 2, 3, 0, 1, 1, 1, 1, 1, 2, 0, 3, 1, 1, 0, 5, 2, 1, 1, 0, 4, 2, 1, 1, 2, 0, 3

Offset: 0

Antti Karttunen, Nov 10 2013

nonn

This sequence relates to the factorial base representation (A007623) in the same way as A213725 relates to the binary system.

			From 11 sprouts the following finite side-tree of "factorial beanstalk":
    18  19
     \  /
  14  15
   \  /
    11
Its leaves are the numbers 14, 18 and 19 (which all occur in A219658), whose factorial base representations (see A007623) are '210', '300' and '301' respectively. The corresponding parent nodes are obtained by subtracting the sum of factorial base digits, thus we get 18-3 = 15 and also 19-4 = 15, thus 15 ('211' in factorial base) is the parent of 18 and 19. For 14 and 15 we get 14-3 = 15-4 = 11, thus 11 is the parent of both 14 and 15, and the common ancestor of all the numbers 11, 14, 15, 18 and 19.
For numbers not occurring in A219666 this sequence gives one more than the maximum number of steps to reach the most distant leaf in such subtrees. In the above case, there is from 11 only one step to 14, but two steps to both 18 and 19. Thus a(11)=2+1=3. For leaves the result is always 1, for example, a(14) = a(18) = a(19) = 1.

Antti Karttunen, Table of n, a(n) for n = 0..10080

Differs from A230426 for the first time at n=34, where a(n)=3, while A230426(34)=4. Cf. also A230427.

A219658 gives the position of ones in this sequence (which are the leaves of the tree).

If A230412(n)=0, a(n)=1; otherwise, if n is in A219666, a(n)=0; otherwise a(n) = 1+max(a(A230423(n)), a(A230424(n))).

A257676 Permutation of nonnegative integers obtained by traversing the tendrils (finite side-trees) of the binary beanstalk in depth-first order, with also each number in the infinite trunk visited, but only after its sister branch has been traversed first.

Original entry on oeis.org

0, 1, 2, 3, 5, 4, 6, 7, 9, 8, 10, 12, 13, 11, 14, 15, 17, 16, 18, 20, 21, 19, 22, 24, 25, 28, 29, 23, 27, 26, 30, 31, 33, 32, 34, 36, 37, 35, 38, 40, 41, 44, 45, 39, 43, 42, 47, 50, 54, 58, 59, 55, 51, 46, 48, 49, 52, 53, 56, 60, 61, 57, 62, 63, 65, 64, 66, 68, 69, 67, 70, 72, 73, 76, 77, 71, 75, 74, 79, 82, 86, 90, 91, 87, 83, 78, 80, 81

Offset: 0

Antti Karttunen, May 04 2015

nonn

			Please look at Paul Tek's illustration: We start at root, 0, go up to 1, visit its left child 2 (which is a leaf), before proceeding the infinite trunk (A179016) to 3, then visit first the leaf 5 at the right hand side, before proceeding the infinite trunk to 4, then visit the leaf 6 at the left hand side, before proceeding the infinite trunk right to 7, from which we first visit the leaf 9 at the right hand side, before proceeding the infinite trunk to 8 at the left hand side. Thus we have ten initial terms of the sequence: 0, 1, 2, 3, 5, 4, 6, 7, 9, 8, ...
From 8 we proceed first to the left 10, because it is not a part of the infinite trunk, and we traverse a finite side-tree ("tendril") of three nodes in order 10, 12, 13, only after which we proceed the infinite trunk to the right, to 11, thus we have the next four terms of the sequence 10, 12, 13, 11.

Inverse: A257677.
Fixed points: A257678.
Cf. A011371, A055938, A179016, A213719, A213723, A213724, A213725, A213727, A213730.
Cf. also A218252.

a(0) = 0; a(1) = 1;
  otherwise set prev = a(n-1);
  if A213719(prev) = 1 [prev is one of the terms in A179016]
   then if A213719(A213723(prev)) = 0, a(n) = A213723(prev),
        else a(n) = A213724(prev);
  else if(A213723(prev) > 0), a(n) = A213723(prev),
  else if(A213724(prev) > 0), a(n) = A213724(prev),
  otherwise,
    a(n) = {the first unvisited node of binary beanstalk tree found when we backtrack out of a finite branch just traversed in depth-first order}.
Other identities and observations:
If a(n-1) is an even term of A055938 then a(n) = a(n-1)+1.

cp's OEIS Frontend

A230425 a(n)=0 if n is in the infinite trunk of factorial beanstalk (in A219666), otherwise 1 + number of steps to reach the farthest leaf in that finite branch of the beanstalk.

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