cp's OEIS Frontend

These are triples matching the pattern (2,1,2), (3,1,2), or (2,1,3).

A sequence of integers is unimodal if it is the concatenation of a weakly increasing and a weakly decreasing sequence.

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.

Triangular numbers with exactly three ones in their binary representation: A000120(a(n)) = 3; triangular numbers in A014311; no more terms less than A014311(1000000) = 6129982174881536975083663305496791903885182827960991744.

Also Heinz numbers of integer partitions with 3 parts, all of which are odd and > 1. These partitions are counted by A001399.

This sequence is inspired by the 3rd problem, proposed by Romania, during the 35th International Mathematical Olympiad in 1994 at Hong Kong (see the link IMO).

This sequence is increasing because there are only these two possibilities:

-> a(n+1) - a(n) = 1 if n has exactly two 1's in its binary representation (A018900);

-> a(n+1) - a(n) = 0 otherwise.

Consequence, for any positive integer m, a(x) = m has at least one solution (answer to the 1st Olympiad question).

Only when m = k*(k-1)/2 + 1 with k >= 2 (A000124 \ {1}), there exists only one n such that a(n) = m, and then n = 2^k+2 where k >= 2 (A052548 \ {3, 4}) (answer to the 2nd Olympiad question).

There are O(log^4 x) members of this sequence up to x. - Charles R Greathouse IV, Mar 29 2013

Intersection of A014311 and A023694.

All terms are odd, since n == A062756(n) (mod 2).

It is likely that a(136) = 1099528404993 is the last term. The next term, if any, is greater than 10^200.

In other words, if k numbers having weight w have occurred, the most recent being a(n-1), then a(n) = k*w. Consequently every integer m > 1 appears A000005(m) times. Whilst there are > 2 ways composite number m may appear, there are only two ways for prime p. The first is consequent to the p_th occurrence of a power of 2. The final appearance of any number m is consequent to the first term in the sequence whose weight is m. For this reason final occurrences are very much delayed.

1 appears twice since A000120(1) = 1, the only fixed point in A000120.

First occurrences of primes are in natural order.

It appears that a(n) <= n, with equality at fixed points 1, 26, 28, ...

The plots have a curious net-like structure.

From Michael De Vlieger, Dec 29 2022: (Start)

a(186) = 188, and for n <= 2^20, there are 694462 occasions of a(n) > n.

Let w(n) = A000120(n) and let c_w(k) be the number of k in this sequence with binary weight w(k). Then this sequence consists of the recursive mapping of f(n) = w(a(n-1)) * c_w(a(n-1)).

Since f(n) is a product of 2 positive numbers, a(n) is odd iff both w(a(n-1)) and c_w(a(n-1)) are odd.

Let S_m = { k : w(k) = m }, thus, S_0 = {0}, S_1 = A000079, S_2 = A018900, S_3 = A014311, etc., with least element (2^m)-1 for m > 0.

Let trajectory T_m comprise a(j) such that w(a(j)) = m. Then a(j) is in S_m.

If the a(j) in T_m appear in order of j, then T_m(1) is such that c_m = 1, T_m(2) is such that c_m = 2, and generally c_m(T_m(k)) = k.

This sequence is composed of trajectories T_m evident in scatterplot. (End)

A067576 describes the sequences with a fixed number of binary bits using antidiagonals.