cp's OEIS Frontend

Complement of the parity sequence of iterated triangular numbers (A117872).

Number of nonisomorphic complete binary trees with leaves colored using two colors. - Brendan McKay, Feb 01 2001

With a(0) = 2, a(n+1) is the number of possible distinct sums between any number of elements in {1,...,a(n)}. - Derek Orr, Dec 13 2014

a(n) = sum of digits of A000462(n). - Reinhard Zumkeller, Mar 27 2011

The length of (number of moves in) Simon Norton's game in A006019 starting with an initial heap of n if both players always take, never put. - R. J. Mathar, May 13 2016

The next two terms, a(10) and a(11), have 111 and 221 digits. - Harvey P. Dale, Jun 10 2011

Interpretation through plane geometry: Start with the a(n)-sided regular polygon, connect all the vertices to create a figure having a(n+1)=A000217(a(n)-1) edges. Repeat to obtain this sequence. - T. D. Noe, May 13 2016

Let y(1) = x1+x2+x3+x4, and define y(n+1) as the plethysm e2[y(n)], where e2 represents the second elementary symmetric function. Then a(n) is y(n) evaluated at x1=x2=x3=x4=1. - Per W. Alexandersson, Jun 06 2020

Each term is the number of coordinate planes in Euclidean space of the dimensionality of the previous term. - Shel Kaphan, Feb 06 2023

a(n) is also the number of maximally balanced trees with n leaves according to the Sackin index.

From Omar E. Pol, Feb 02 2018: (Start)

Also the sequence can be written as an irregular triangle read by rows in which the row lengths are the terms of A011782 (see example section).

Column 1 gives A000012. Column 2 gives A000027.

Conjecture 1: column 3 gives the nonzero terms of A000217.

Conjecture 2: column 4 gives the nonzero terms of A000330. (End)

Comment from the author, Feb 02 2018: Denote the (i,j)-th entry of the irregular triangle described above (row i, column j) with T(i,j), i >= 1 and 1 <= j <= A011782(i-1). Then, rows 1 and 2 contain a 1 each (and they denote the number of trees minimizing the Sackin index with 1 and two leaves, respectively) and row i ranges from 2^(i-2) + 1 to 2^(i-1) leaves and for each of these numbers gives the number of trees with minimum Sackin index. E.g., row 4 gives the number of such trees for 2^(4-2) + 1 = 5 leaves up to 2^(4-1) = 8 leaves.

From Omar E. Pol, Mar 01 2018: (Start)

Conjecture 3: column 5 gives the nonzero terms of A212415.

Conjecture 4: row sums give 1 together with A006893.

Conjecture 5: T(i,j) has i >= 0 where "i" is the height of the rooted trees.

Conjecture 6: For i >= 1, j is the number of leaves minus 2^(i-1). (End)

The first row with k columns is the A006893(k)-th. The last row with k columns comprises the first k terms of A006893.

Fermat asserted, and Gauss proved, that every number is the sum of three triangular numbers (cf. A002636).

For exactly half of the integers k in 1..11898, decomposing k into triangular numbers by the greedy algorithm (i.e., letting T1 be the largest triangular number <= k, then letting T2 be the largest triangular number <= k-T1, etc.) yields a decomposition of k into three or fewer positive triangular numbers, but for any K > 11898, the greedy algorithm decomposes more than half of the integers k in 1..K into four or more positive triangular numbers.

Even an approach that assigns only T1 "greedily" but allows T2 to be any triangular number will usually not yield a set of three triangular numbers whose sum is k: for more than half of the integers k in 1..K for any K > 40304762, no such sum exists in which the largest of the three triangular numbers is the largest triangular number <= k. The smallest such k is a(1)=20 (see Example section).

For some values of k, there exists no set of three triangular numbers summing to k unless the largest of the three is neither T(m(k)) nor T(m(k)-1); the smallest of these is a(2)=50, for which a solution to T(m(k)-2) + T2 + T3 = k does exist (see Example section).