cp's OEIS Frontend

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

A composition of n is a finite sequence of positive integers with sum n.

For the operation of taking the sequence of run-lengths of a finite sequence, runs-resistance is defined as the number of applications required to reach a singleton.

Except for the k = 0 column and the n = 0 and n = 1 rows, this is the triangle appearing on page 3 of Lenormand, which is A319411. Unlike A318928, we do not here require that a(n) >= 1.

The n = 0 row is chosen to ensure that the row-sums are A000079, although the empty word arguably has indeterminate runs-resistance.

For the Stolarsky representation of n, see the C. Mongoven link.

Conjecture: a(n) is the length of row n-1 of A385886. To obtain it, first take maximal anti-run lengths of binary indices of each nonnegative integer (giving A384877), then remove all duplicate rows (giving A385886), and finally take the length of each remaining row. For sum instead of length we appear to have A200648. For runs minus 1 instead of anti-runs see A200650. - Gus Wiseman, Jul 21 2025

For the Stolarsky representation of n, see the C. Mongoven link.

Conjecture: a(n) is the sum of row n-1 of A385886. To obtain it, first take maximal anti-run lengths of binary indices of each nonnegative integer (giving A384877), then remove all duplicate rows (giving A385886), and finally take the sum of each remaining row. For length instead of sum we appear to have A200649. - Gus Wiseman, Jul 21 2025

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

For the Stolarsky representation of n, see the C. Mongoven link.

a(n+1), n >= 1, gives the size of the n-th generation of each of the "[male-female] pair of Fibonacci rabbits" in the Fibonacci rabbits tree read right-to-left by row, the first pair (the root) being the 0th generation. (Cf. OEIS Wiki link below.) - Daniel Forgues, May 07 2015

From Daniel Forgues, May 07 2015: (Start)

Concatenation of:

0: 1,

1: 0,

2: 0,

3: 1, 0,

4: 1, 1, 0,

5: 2, 1, 1, 1, 0,

6: 2, 2, 1, 2, 1, 1, 1, 0,

(...),

where row n, n >= 3, is row n-1 prepended by incremented row n-2. (End)

For n >= 3, this algorithm yields the next F_n terms of the sequence, where F_n is the n-th Fibonacci number (A000045). Since it is asymptotic to (phi^n)/sqrt(5), the number of terms thus obtained grows exponentially at each step! - Daniel Forgues, May 22 2015

Conjecture: a(n) is one less than the length of row n-1 of A385817. To obtain it, first take maximal run lengths of binary indices of each nonnegative integer (giving A245563), then remove all duplicate rows (giving A385817), and finally take the length of each remaining row and subtract 1. For sum instead of length we appear to have A200648. For anti-runs instead of runs we appear to have A341259 = A200649-1. - Gus Wiseman, Jul 21 2025

How is this related to A117479? - R. J. Mathar, Aug 10 2025

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

If the k-th composition in standard order is y, then the standard composition number of y is defined to be k.

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

This is the triangle A384877, except all duplicates after the first instance of each composition are removed. It lists all compositions in order of their first appearance as a row of A384877.

Riordan array (x/(1-x)^2, x^2/(1-x)^2).

Mirror image of triangle in A119900.

A203322*A130595 as infinite lower triangular matrices. - Philippe Deléham, Jan 05 2011

From Gus Wiseman, Jul 07 2025: (Start)

Also the number of subsets of {1..n} containing n with k maximal runs (sequences of consecutive elements increasing by 1). For example, row n = 5 counts the following subsets:

{5} {1,5} {1,3,5}

{4,5} {2,5}

{3,4,5} {3,5}

{2,3,4,5} {1,2,5}

{1,2,3,4,5} {1,4,5}

{2,3,5}

{2,4,5}

{1,2,3,5}

{1,2,4,5}

{1,3,4,5}

For anti-runs instead of runs we have A053538.

Without requiring n see A210039, A202023, reverse A098158, A109446.

(End)

This is the number k such that the k-th composition in standard order is the reversed sequence of lengths of the maximal anti-runs of the binary indices of n. Here, the binary indices of n are row n of A048793, and the k-th composition in standard order is row k of A066099. For example, the binary indices of 100 are {3,6,7}, with maximal anti-runs ((3,6),(7)), with reversed lengths (1,2), which is the 6th composition in standard order, so a(100) = 6. - Gus Wiseman, Jul 27 2025