A127971 -id:A127971 - cp's OEIS frontend

A127970 Number triangle A127967 modulo 2.

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

Offset: 0

Paul Barry, Feb 09 2007

Row sums are A127971.

			Triangle begins
1,
1, 1,
1, 0, 1,
1, 0, 0, 1,
1, 0, 1, 0, 1,
1, 0, 0, 0, 1, 1,
1, 0, 1, 0, 0, 0, 1,
1, 0, 0, 0, 0, 0, 0, 1,
1, 0, 1, 0, 1, 0, 0, 0, 1,
1, 0, 0, 0, 1, 0, 0, 0, 1, 1,
1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1,
1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1,
1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1

A131217 Triangle read by rows: T(n, k) = A047999(n, k) + (n+1 mod 2)*[k=1].

Original entry on oeis.org

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

Offset: 0

Roger L. Bagula, Sep 27 2007

An XOR of the sequence terms of A047999 is the algorithm.

Former name: Triangular sequence of a Gray code type made from Pascal's triangle modulo 2 as: T(n, k) = (b(n, k) + b(n, k+1)) mod 2, where b(n, k) = A047999(n, k).

			Triangle begins as:
  1;
  1, 1;
  1, 1, 1;
  1, 1, 1, 1;
  1, 1, 0, 0, 1;
  1, 1, 0, 0, 1, 1;
  1, 1, 1, 0, 1, 0, 1;
  1, 1, 1, 1, 1, 1, 1, 1;
  1, 1, 0, 0, 0, 0, 0, 0, 1;
  1, 1, 0, 0, 0, 0, 0, 0, 1, 1;
  1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A001316, A047999, A122944, A127971.

A047999:= func< n, k | BitwiseAnd(n-k, k) eq 0 select 1 else 0 >;
A131217:= func< n,k | k ne 1 select A047999(n,k) else 1 >;
[A131217(n, k): k in [0..n], n in [0..15]]; // G. C. Greubel, Sep 03 2025

(* First program *)
A047999[n_, k_] := Mod[Binomial[n, k], 2];
T[n_, k_]:= Mod[A047999[n-1,k-1] + A047999[n-1,k], 2] + Boole[n==0] + Mod[n+1, 2]*Boole[k==1];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten
(* Second program *)
Table[Boole[BitAnd[n-k,k]==0] +Mod[n+1,2]*Boole[k==1], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 03 2025 *)

def A047999(n,k): return int(not ~n & k)
def A131217(n,k): return A047999(n,k) + ((n+1)%2)*int(k==1)
flatten([[A131217(n,k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Sep 03 2025

T(n, k) = (A047999(n-1, k-1) + A047999(n-1, k)) mod 2 + [n=0] + (n+1 mod 2)*[k=1].
From G. C. Greubel, Sep 03 2025: (Start)
T(n, k) = A047999(n, k) + (n+1 mod 2)*[k=1].
Sum_{k=0..n} T(n, k) = A001316(n) + (1 + (-1)^n)/2 - [n=0].
Sum_{k=0..n} (-1)^k*T(n, k) = (1 + (-1)^n)/2*(A001316(n/2) - 1) + [n=0].
Sum_{k=0..floor(n/2)} T(n-k, k) = A127971(n) - [n=1]. (End)

Edited by G. C. Greubel, Sep 03 2025

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A127970 Number triangle A127967 modulo 2.

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A131217 Triangle read by rows: T(n, k) = A047999(n, k) + (n+1 mod 2)*[k=1].

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