A375972 a(n) = binomial(n,k(n)), where k(2) = 1, k(n) = k(n-1) + (a(n-1) mod 2).

2, 3, 6, 10, 15, 35, 70, 126, 210, 330, 495, 1287, 3003, 6435, 12870, 24310, 43758, 75582, 125970, 203490, 319770, 490314, 735471, 2042975, 5311735, 13037895, 30421755, 67863915, 145422675, 300540195, 601080390, 1166803110, 2203961430, 4059928950, 7307872110, 12875774670, 22239974430

Offset: 2

Author

Walter Robinson, Sep 04 2024

Keywords

• nonn

Comments

A way to travel down the Pascal's triangle starting in the center of the third row, where you turn left at every even number, and turn right at every odd number, while descending a row at every step.

Pascal's triangle is symmetric, thus switching the turn directions would not change the sequence.

Examples

			a(3) = 3, and k(3) = 1.
To derive a(4), we first find that k(4) = k(3) + (a(3) mod 2) = 1 + (3 mod 2) = 2.
Therefore a(4) = binomial(4,k(4)) = binomial(4,2) = 6.
		

Links

• Robert Israel, Table of n, a(n) for n = 2..3418

Crossrefs

Cf. A007318.

Initial conditions of n=0, k(0)=0 would generate A000012.

Programs

Maple

k[2]:= 1:
for n from 2 to 50 do
  a[n]:= binomial(n, k[n]);
  k[n+1]:= k[n] + (a[n] mod 2);
od:
seq(a[n],n=2..50); # Robert Israel, Jan 27 2025

Mathematica

Module[{k}, FoldList[{Binomial[#2, k = #[[2]] + Mod[#[[1]], 2]], k} &, {2, 1}, Range[3, 50]][[All, 1]]] (* Paolo Xausa, Jan 28 2025 *)

PARI

lista(nn) = my(va=vector(nn), vk=vector(nn)); vk[2] = 1; va[2] = binomial(2, vk[2]); for (n=3, nn, vk[n] = vk[n-1] + va[n-1] % 2; va[n] = binomial(n, vk[n]);); vector(nn-1, k, va[k+1]); \\ Michel Marcus, Sep 27 2024

Python

def genNextRow(arr):
    nextRow = []
    nextRow.append(arr[0])
    for i in range(1, len(arr)):
        nextRow.append(arr[i-1]+arr[i])
    nextRow.append(arr[len(arr)-1])
    return nextRow
pascal = [[1],[1,1]]
n = 0
index = 1
while n < 30:
    pascal.append(genNextRow(pascal[1]))
    pascal.pop(0)
    print(pascal[1][index])
    index = index + (pascal[1][index] % 2)
    n += 1

Extensions

More terms from Michel Marcus, Sep 30 2024