cp's OEIS Frontend

It seems that for all n, a(A000079(n)) = A003945(n).

a(0)=0; thereafter if n occurs as a term of A055938, a(n)=its position in A055938, otherwise zero. This works as an "inverse" function for A055938 in a sense that a(A055938(n)) = n for all n.

a(n)*A213714(n) = 0 for all n.

For all n, a(A000079(n)) = A000225(n+1), i.e. a(2^n) = (2^(n+1))-1.

For n>=1, a(A000225(n)) = A000325(n).

This permutation is obtained by "entangling" even and odd numbers with complementary pair A005187 & A055938, meaning that it can be viewed as a binary tree. Each child to the left is obtained by applying A005187(n+1) to the parent node containing n, and each child to the right is obtained as A055938(n):

0

|

...................1...................

3 2

7......../ \........6 4......../ \........5

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

15 14 11 13 8 9 10 12

31 30 26 29 22 24 25 28 16 17 18 20 19 21 23 27

etc.

For n >= 1, A256991(n) gives the contents of the immediate parent node of the node containing n, while A070939(n) gives the total distance to 0 from the node containing n, with A256478(n) telling how many of the terms encountered on that journey are terms of A005187 (including the penultimate 1 but not the final 0 in the count), while A256479(n) tells how many of them are terms of A055938.

Permutation A233278 gives the mirror image of the same tree.

This permutation is obtained by "entangling" even and odd numbers with complementary pair A055938 & A005187, meaning that it can be viewed as a binary tree. Each child to the left is obtained by applying A055938(n) to the parent node containing n, and each child to the right is obtained as A005187(n+1):

0

|

...................1...................

2 3

5......../ \........4 6......../ \........7

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

12 10 9 8 13 11 14 15

27 23 21 19 20 18 17 16 28 25 24 22 29 26 30 31

etc.

For n >= 1, A256991(n) gives the contents of the immediate parent node of the node containing n, while A070939(n) gives the total distance to zero at the root from the node containing n, with A256478(n) telling how many of the terms encountered on that journey are terms of A005187 (including the penultimate 1 but not the final 0 in the count), while A256479(n) tells how many of them are terms of A055938.

Permutation A233276 gives the mirror image of the same tree.

The array is read by antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.

This is transpose of array A256995.

If we assume that a(1) = 1 (but which is not explicitly included here because outside of the array proper), then A256998 gives the inverse permutation.

In other words, if n = A005187(k) for some k >= 1, then a(n) = k, otherwise it must be that n = A055938(h) for some h, and then a(n) = h.

Each n occurs exactly twice, first at a(A005187(n)), then at a(A055938(n)). Cf. also A257126.

When iterating a(n), a(a(n)), a(a(a(n))), etc, A256993(n) gives the number of steps to reach one, from any starting value n >= 1.

This sequence is one instance of Clark Kimberling's generic dispersion arrays. Paraphrasing his explanation in A191450, mutatis mutandis, we have the following definition:

Suppose that s is an increasing sequence of positive integers, that the complement t of s is infinite, and that t(1)=1. The dispersion of s is the array D whose n-th row is (t(n), s(t(n)), s(s(t(n))), s(s(s(t(n)))), ...). Every positive integer occurs exactly once in D, so that, as a sequence, D is a permutation of the positive integers. The sequence u given by u(n) = {index of the row of D that contains n} is a fractal sequence. In this case s(n) = A055938(n), t(n) = A005187(n) [from term A005187(1) onward] and u(n) = A254112(n).

For other examples of such sequences, see the Crossrefs section. For a general introduction, please follow the Kimberling references.

The main diagonal: 1, 6, 20, 60, 154, 377, 887, 2040, 4598, 10229, 22515, 49139, ...

This is dispersion array A254105 transposed. Please see there for a description.

The array is read by antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.

Provided that I understand Kimberling's terminology correctly, this array is the dispersion of sequence b(n) = A005187(n+1), for n>=1: A005187[2..] = [3, 4, 7, 8, 10, 11, ...]. The left column is the complement of that sequence, which is {1} followed by A055938. - Antti Karttunen, Apr 17 2015