A380231 - OEIS

A380231

Alternating row sums of triangle A237591.

1, 2, 1, 2, 1, 4, 3, 4, 5, 4, 3, 6, 5, 4, 7, 8, 7, 8, 7, 10, 9, 8, 7, 10, 11, 10, 9, 12, 11, 14, 13, 14, 13, 12, 15, 16, 15, 14, 13, 16, 15, 18, 17, 16, 19, 18, 17, 20, 21, 22, 21, 20, 19, 22, 21, 24, 23, 22, 21, 24, 23, 22, 25, 26, 25, 28, 27, 26, 25, 28, 27, 32, 31, 30, 29, 28, 31, 30, 29

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

Consider the symmetric Dyck path in the first quadrant of the square grid described in the n-th row of A237593. Let C = (A240542(n), A240542(n)) be the middle point of the Dyck path.

a(n) is also the coordinate on the x axis of the point (a(n),n) and also the coordinate on the y axis of the point (n,a(n)) such that the middle point of the line segment [(a(n),n),(n,a(n))] coincides with the middle point C of the symmetric Dyck path.

The three line segments [(a(n),n),C], [(n,a(n)),C] and [(n,n),C] have the same length.

For n > 2 the points (n,n), C and (a(n),n) are the vertices of a virtual isosceles right triangle.

For n > 2 the points (n,n), C and (n,a(n)) are the vertices of a virtual isosceles right triangle.

For n > 2 the points (a(n),n), (n,n) and (n,a(n)) are the vertices of a virtual isosceles right triangle.

LINKS

Paolo Xausa, Table of n, a(n) for n = 1..10000

FORMULA

a(n) = 2*A240542(n) - n.

a(n) = n - 2*A322141(n).

a(n) = A240542(n) - A322141(n).

EXAMPLE

For n = 14 the 14th row of A237591 is [8, 3, 1, 2] hence the alternating row sum is 8 - 3 + 1 - 2 = 4, so a(14) = 4.

On the other hand the 14th row of A237593 is the 14th row of A237591 together with the 14 th row of A237591 in reverse order as follows: [8, 3, 1, 2, 2, 1, 3, 8].

Then with the terms of the 14th row of A237593 we can draw a Dyck path in the first quadrant of the square grid as shown below:

(y axis)

. (4,14) (14,14)

._ _ _ . _ _ _ _ .

. |

. |_

. |

. |_ _

. C |_ _ _

. |

. . (14,4)

. |