cp's OEIS Frontend

T differs from A212162 first at (n,k) = (5,10): T(5,10) = -3101089710, A212162(5,10) = -3101089711.

The staggered hexagonal square grid graph SH_(n,n) has n^2 = A000290(n) vertices and (n-1)*(3*n-1) = A045944(n-1) edges. The chromatic polynomial of SH_(n,n) has n^2+1 = A002522(n) coefficients.

In general, column k>=2 is asymptotic to k^(n*(n-1)) / (n!*(n-1)!). - Vaclav Kotesovec, Jun 05 2015

Similar to A274528, but the triangle here is a right triangle.

The same rule applied to an equilateral triangle gives A274528.

By analogy, the offset for both rows and columns is the same as the offset of A274528.

We construct the triangle by reading from left to right in each row, starting with T(0,0) = 0.

Presumably every diagonal and every column is also a permutation of the nonnegative integers, but the proof does not seem so straightforward. Of course neither the rows nor the antidiagonals are permutations of the nonnegative integers, since they are finite in length.

Omar E. Pol's conjecture that every column and every diagonal of the triangle is a permutation of the nonnegative integers is true: see the link. - N. J. A. Sloane, Jun 07 2017

It appears that the numbers generally appear for the first time in or near the right border of the triangle.

Theorem 1: the middle diagonal gives A000004 (the zero sequence).

Proof: T(0,0) = 0 by definition. For the next rows we have that in row 1 there are no zeros because the first term belongs to a column that contains a zero and the second term belongs to a diagonal that contains a zero. In row 2 the unique zero is T(2,1) = 0 because the preceding term belongs to a column that contains a zero and the following term belongs to a diagonal that contains a zero. Then we have two recurrences for all rows of the triangle:

a) If T(n,k) = 0 then row n+1 does not contain a zero because every term belongs to a column that contains a zero or it belongs to a diagonal that contains a zero.

b) If T(n,k) = 0 the next zero is T(n+2,k+1) because every preceding term in row n+2 is a positive integer because it belongs to a column that contains a zero and, on the other hand, the column, the diagonal and the antidiagonal of T(n+2,k+1) do not contain zeros.

Finally, since both T(n,k) = 0 and T(n+2,k+1) = 0 are located in the middle diagonal, the other terms of the middle diagonal are zeros, or in other words: the middle diagonal gives A000004 (the zero sequence). QED

Theorem 2: all zeros are in the middle diagonal.

Proof: consider the first n rows of the triangle. Every element located above or at the right-hand side of the middle diagonal must be positive because it belongs to a diagonal that contains one of the zeros of the middle diagonal. On the other hand every element located below the middle diagonal must be positive because it belongs to a column that contains one of the zeros of the middle diagonal, hence there are no zeros outside of the middle diagonal, or in other words: all zeros are in the middle diagonal. QED

From Hartmut F. W. Hoft, Jun 12 2017: (Start)

T(2k,k) = 0, for all k >= 0, and T(n,{(n-1)/2,(n+2)/2,(n-2)/2,(n+1)/2}) = 1, for all n >= 0 with n mod 8 = {1,2,4,5} respectively, and no 0's or 1's occur in other positions. The triangle of positions of 0's and 1's for this sequence is the triangle in the Comment section of A274651 with row and column indices and values shifted down by one.

The sequence of rows containing 1's is A047613 (n mod 8 = {1,2,4,5}), those containing only 1's is A016813 (n mod 8 = {1,5}), those containing both 0's and 1's is A047463 (n mod 8 = {2,4}), those containing only 0's is A047451 (n mod 8 = {0,6}), and those containing neither 0's nor 2's is A004767 (n mod 8 = {3,7}).

(End)

Indexed as a square array A(n,k): If X is an (n+k)-set and Y a fixed k-subset of X then A(n,k) is equal to the number of n-subsets of X intersecting Y. - Peter Luschny, Apr 20 2012

Also, number of paths along a corridor with width k, starting from one side (from H. Bottomley's comment in A061551).

Rows converge to binomial(n,floor(n/2)) (A001405).

Note that the rows and columns start at 1, which for example obscures the fact that the first row refers to A000007 and not to A000004. A better choice is the indexing 0 <= k and 0 <= n. The Maple program below uses this indexing and builds only on the roots of -1. - Peter Luschny, Sep 17 2020

If the sequence ends with (1,1,0) Abel wins; if it ends with (1,0,0) Kain wins.

Abel(n) = A002620(n-1) = (2*n*(n - 2) + 1 - (-1)^n)/8.

Kain(n) = A004526(n-1) = floor((n - 1)/2).

Win probability for Abel = sum(Abel(n)/2^n) = 2/3.

Win probability for Kain = sum(Kain(n)/2^n) = 1/3.

Mean length of the game = sum(n*a(n)/2^n) = 16/3.

Essentially the same as A035106. - R. J. Mathar, Oct 27 2011

The sequence 2*a(n) is denoted as chi(n) by McKee (1994) and is the degree of the division polynomial f_n as a polynomial in x. He notes that "If x is given weight 1, a is given weight 2, and b is given weight 3, then all the terms in f_n(a, b, x) have weight chi(n)". - Michael Somos, Jan 09 2015

In Duistermaat (2010), at the end of section 11.2 The Elliptic Billiard, on page 492 the number of k-periodic fibers counted with multiplicities of the QRT root is given by equation (11.2.8) as "1/4 k^2 + 3{k/2}(1 - {k/2}) - 1 = n^2 - 1 when k = 2n, n^2 + n when k = 2n+1, for every integer k." - Michael Somos, Mar 14 2023

T(n,k) = T(n+k,-k).