A117963 -id:A117963 - cp's OEIS frontend

A117964 a(n) = A117963(n) mod 2.

1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Paul Barry, Apr 05 2006

a(3n+2) = 0, a(3n) = a(3n+1). a(3n) may be equal to A088917(n).

Antti Karttunen, Table of n, a(n) for n = 0..16384
Index entries for characteristic functions

Cf. A000035, A005836, A117963, A081601, A088917.

a(n)=sum{k=0..floor(n/2), L(C(n-k,k)/3)} mod 2 where L(j/p) is the Legendre symbol of j and p.

a(2*A081601(n)) = a(1+2*A081601(n)) = 1. [Conjectured, also these two formulas together seem to give the positions of all 1's] - Antti Karttunen, Jan 01 2023

A117947 T(n,k)=L(C(n,k)/3) where L(j/p) is the Legendre symbol of j and p.

Original entry on oeis.org

1, 1, 1, 1, -1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, -1, 1, 1, -1, 1, 1, 0, 0, -1, 0, 0, 1, 1, 1, 0, -1, -1, 0, 1, 1, 1, -1, 1, -1, 1, -1, 1, -1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, -1, 1, 0, 0, 0, 0, 0, 0, 1, -1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, -1, 1, 1, -1, 1, 0, 0, 0, 1, -1, 1, 1, -1, 1

Offset: 0

Paul Barry, Apr 05 2006

Row sums are A059126. Diagonal sums are A117963. Could be called the Legendre-binomial matrix for p=3.

The matrix square equals triangle A117939; the matrix log equals triangle A120854 divided by 2. - Paul D. Hanna, Jul 08 2006

			Triangle begins:
  1;
  1, 1;
  1, -1, 1;
  1, 0, 0, 1;
  1, 1, 0, 1, 1;
  ...

Cf. A117939 (matrix square), A120854 (2*log).

T(n,k)=(binomial(n,k)+1)%3-1 - Paul D. Hanna, Jul 08 2006

T(n,k) = balanced ternary digit of C(n,k) mod 3. - Paul D. Hanna, Jul 08 2006

cp's OEIS Frontend

A117964 a(n) = A117963(n) mod 2.

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