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%I #22 May 24 2023 07:33:04
%S 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,
%T 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,
%U 1,0,1,1,0,1,1,1,0,1,1,0,1,1,1,0,1,1
%N a(n) = number of squares in the n-th antidiagonal of the natural number array, A000027.
%C The positions of 1 (i.e., indices of antidiagonals that contain a square) are A022846.
%C Deleting the initial 1 leaves the difference sequence of A061288.
%F a(n) = 1 if A064784(n) < n, and a(n) = 0 otherwise. - _Kevin Ryde_, May 24 2023
%e The natural number array has this corner:
%e 1, 2, 4, 7, 11, 16, 22, 29, ...
%e 3, 5, 8, 12, 17, 23, 30, 38, ...
%e 6, 9, 13, 18, 24, 31, 39, 48, ...
%e 10, 14, 19, 25, 32, 40, 49, 59, ...
%e 15, 20, 26, 33, 41, 50, 60, 71, ...
%e 21, 27, 34, 42, 51, 61, 72, 84, ...
%e in which the squares 1,4,9,16,25, are in these antidiagonals:
%e (1), (4,5,6), (7,8,9,10), (16,17,18,19,20,21).
%p [seq](floor(sqrt(n*(n+1)/2))-floor(sqrt((n-1)*n/2)),n=1..200); # _Robert Israel_, Mar 14 2023
%t r[n_] := Range[n (n - 1)/2 + 1, n (n + 1)/2];
%t t = Table[Intersection[r[n], Range[120]^2], {n, 1, 120}];
%t Map[Length, t]
%o (PARI) a(n) = my(r); sqrtint((n^2+n)>>1,&r); r<n; \\ _Kevin Ryde_, May 24 2023
%Y Cf. A000027, A000290, A061288, A064784.
%Y Cf. A022846 (indices of 1's), A063957 (indices of 0's).
%K nonn,easy
%O 1
%A _Clark Kimberling_, Mar 11 2023
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