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

2,1

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.

Links

Table of n, a(n) for n=2..38.

Example

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.

Prog

(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
(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

Crossrefs

Cf. A007318.

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

Sequence in context: A055789 A238891 A048681 * A051891 A108062 A347117

Adjacent sequences: A375969 A375970 A375971 * A375973 A375974 A375975

Keyword

nonn

Author

Walter Robinson, Sep 04 2024

Extensions

More terms from Michel Marcus, Sep 30 2024

Status

approved