A361433 a(n) = number of squares in the n-th antidiagonal of the natural number array, A000027.

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

Offset

1

Comments

The positions of 1 (i.e., indices of antidiagonals that contain a square) are A022846.

Deleting the initial 1 leaves the difference sequence of A061288.

Links

Table of n, a(n) for n=1..86.

Formula

a(n) = 1 if A064784(n) < n, and a(n) = 0 otherwise. - Kevin Ryde, May 24 2023

Example

The natural number array has this corner:
   1,  2,  4,  7, 11, 16, 22, 29, ...
   3,  5,  8, 12, 17, 23, 30, 38, ...
   6,  9, 13, 18, 24, 31, 39, 48, ...
  10, 14, 19, 25, 32, 40, 49, 59, ...
  15, 20, 26, 33, 41, 50, 60, 71, ...
  21, 27, 34, 42, 51, 61, 72, 84, ...
in which the squares 1,4,9,16,25, are in these antidiagonals:
(1), (4,5,6), (7,8,9,10), (16,17,18,19,20,21).

Maple

[seq](floor(sqrt(n*(n+1)/2))-floor(sqrt((n-1)*n/2)), n=1..200); # Robert Israel, Mar 14 2023

Mathematica

r[n_] := Range[n (n - 1)/2 + 1, n (n + 1)/2];
t = Table[Intersection[r[n], Range[120]^2], {n, 1, 120}];
Map[Length, t]

Prog

(PARI) a(n) = my(r); sqrtint((n^2+n)>>1, &r); r<n; \\ Kevin Ryde, May 24 2023

Crossrefs

Cf. A000027, A000290, A061288, A064784.

Cf. A022846 (indices of 1's), A063957 (indices of 0's).

Sequence in context: A365938 A117567 A307183 * A117568 A093521 A187948

Adjacent sequences: A361430 A361431 A361432 * A361434 A361435 A361436

Keyword

nonn,easy

Author

Clark Kimberling, Mar 11 2023

Status

approved