A265754 Reduced frequency counts for A004001: a(n) = A265332(n+1) - A036987(n).

1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 1, 2, 3, 4, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 5, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 3, 1, 1, 1, 2, 1, 1, 2, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5

Offset: 1

Author

Antti Karttunen, Jan 10 2016

Keywords

• nonn

Comments

Can be generated recursively by first setting R_1 = (1), after which each R_n is obtained by replacing in R_{n-1} each term k with terms 1 .. k, followed by final n. This sequence is then obtained by concatenating all levels R_1, R_2, ..., R_inf together. See page 230 in Kubo-Vakil paper (page 6 in PDF).

Deleting all 1's and decrementing the remaining terms by one gives the sequence back.

Comment from N. J. A. Sloane, Nov 05 2017: (Start)

The following simple Pascal-like triangle produces the same sequence. Construct a triangle T(n,k) of strings (with 0 <= k <= n), where T(0,0) = {1}, T(n,n) = {n+1}, and otherwise T(n,k) is the concatenation of T(n-1,k-1) and T(n-1,k). The first few rows of the triangle (where the strings T(n,k) are shown without spaces for legibility) are:

1

1,2

1,12,3

1,112,123,4

1,1112,112123,1234,5

1,11112,1112112123,1121231234,12345,6

...

Now read the strings across the rows to get the sequence. T(n,k) has length binomial(n,k). (End)

Examples

			Illustration of the sequence as a tree:
             1
            / \
           1   2
          /   /|\
         1   1 2 3_________
        /   / /| | \  \    \
       1   1 1 2 1  2  3__  4________
      /   / / /| | / \ |\ \  \ \ \ \ \
     1   1 1 1 2 1 1 2 1 2 3  1 2 3 4 5
etc.
Compare with the illustration in A265332.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..8191
• T. Kubo and R. Vakil, On Conway's recursive sequence, Discr. Math. 152 (1996), 225-252.

Crossrefs

Cf. A004001, A036987, A054429, A070939, A265332, A266348, A266349.

Cf. A000225 (positions of records, where n appears first time).

Cf. A266640 (obtained from the mirror image of the same tree).

See A293959 for another version.

Formula

a(n) = A265332(n+1) - A036987(n).

As a recurrence: If A036987(n) = 1 [when n is of the form 2^k -1], a(n) = A070939(n), else if a(n+1) = 1, a(n) = a(2^A000523(n) - A266349(n)), otherwise a(n) = a(n+1)-1.

Other identities. For all n >= 1:

a(n) = A266640(A054429(n)).

a(A000225(n)) = n.