cp's OEIS Frontend

From Kolosov Petro, Apr 12 2020: (Start)

Let A(m, r) = A302971(m, r) / A304042(m, r).

Let L(m, n, k) = Sum_{r=0..m} A(m, r) * k^r * (n - k)^r.

Then T(n, k) = L(2, n, k).

Fifth power can be expressed as row sum of triangle T(n, k).

T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)

From Kolosov Petro, Apr 12 2020: (Start)

Let A(m, r) = A302971(m, r) / A304042(m, r).

Let L(m, n, k) = Sum_{r=0..m} A(m, r) * k^r * (n - k)^r.

Then T(n, k) = L(3, n, k).

T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)

Column k, except for the initial k-1 0's, is an arithmetic progression with first term 1 and common difference 2(k-1). Row sums yield A116731. First column of the inverse matrix is A129779.

Studied by Paul Curtz circa 1993.

From Rogério Serôdio, Dec 19 2017: (Start)

T(n, k) gives the number of distinct sums of 2(k-1) elements in {1,1,2,2,...,n-1,n-1}. For example, T(6, 2) = the number of distinct sums of 2 elements in {1,1,2,2,3,3,4,4,5,5}, and because each sum from the smallest 1 + 1 = 2 to the largest 5 + 5 = 10 appears, T(6, 2) = 10 - 1 = 9. [In general: 2*(Sum_{j=1..(k-1)} n-j) - (2*(Sum_{j=1..k-1} j) - 1) = 2*n*(k-1) - 4*(k-1)*k/2 + 1 = 2*(k-1)*(n-k) + 1 = T(n, k). - Wolfdieter Lang, Dec 20 2017]

T(n, k) is the number of lattice points with abscissa x = 2*(k-1) and even ordinate in the closed region bounded by the parabola y = x*(2*(n-1) - x) and the x axis. [That is, (1/2)*y(2*(k-1)) + 1 = T(n, k). - Wolfdieter Lang, Dec 20 2017]

Pascal's triangle (A007318, but with apex in the middle) is formed using the rule South = West + East; the rascal triangle A077028 uses the rule South = (West*East + 1)/North; the present triangle uses a similar rule: South = (West*East + 2)/North. See the formula section for this recurrence. (End)

a(n) is the number of binary words with length <= n+1 which contain at least one 0 and one 1 and have at most one ascent. - Amelia Gibbs, May 21 2024

Antidiagonal sums are given by 2*A055795.

Rows/columns n are binomial transform of {n, A152947(n+1), n, 0, 0, 0, ...}.

Some primes in the array are

n = 1: {2, 7, 11, 29, 37, 67, 79, 137, 191, 211, 277, 379, ...} = A055469, primes of the form k*(k + 1)/2 + 1;

n = 3: {3, 7, 37, 53, 479, 653, 1249, 1619, 2503, 3727, 4349, 5737, 7109, 8179, 9803, 11839, 12107, ...};

n = 4: {11, 37, 79, 137, 211, 821, 991, 1597, 1831, 2081, 2347, ...} = A188382, primes of the form 8*(2*k - 1)^2 + 2*(2*k - 1) + 1.

The rascal triangle (A077028) can be generated by either of the rules South = (East*West+1)/North or South = East+West+1-North; this number triangle can be generated by either of the rules South = (East*West+3)/North or South = East+West+1-North.

It is more suggestive to observe that N*S-E*W = 1 or 3 in the two cases, and (N+S)-(E+W) = 1 in both cases. In fact "3" in the present definition can be replaced by any integer c, and we get a triangle of integers with N*S-E*W = c and (N+S)-(E+W) = 1. I say "suggestive", because these rules also arise in frieze patterns. - N. J. A. Sloane, Aug 28 2019

Row sums are n*(n-1)/2+1 (essentially A000124). Diagonal sums are A114220. First difference triangle of A077028, when this is viewed as a number triangle.

From R. J. Mathar, Mar 22 2013: (Start)

The matrix inverse is

1;

0, 1;

0, -1, 1;

0, 1, -2, 1;

0, -2, 4, -3, 1;

0, 6, -12, 9, -4, 1;

0, -24, 48, -36, 16, -5, 1;

0, 120, -240, 180, -80, 25, -6, 1;

0, -720, 1440, -1080, 480, -150, 36, -7, 1;

... apparently related to A208058. (End)

Number of permutations of length n avoiding simultaneously the patterns 132 and 321 with k left-to-right maxima (resp., right-to-left minima). A left-to-right maximum (resp., right-to-left minimum) in a permutation p(1)p(2)...p(n) is a position i such that p(j) < p(i) for all j < i (resp., p(j) > p(i) for all j > i). - Sergey Kitaev, Nov 18 2023

The rascal triangle (A077028) can be generated by South = (East*West+1)/North or South = East+West+1-North; this triangle can be generated by South = (East*West+1)/North, South = East+West+2-North.