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A361433
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a(n) = number of squares in the n-th antidiagonal of the natural number array, A000027.
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0
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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
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OFFSET
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1
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COMMENTS
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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.
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LINKS
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FORMULA
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EXAMPLE
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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).
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MAPLE
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[seq](floor(sqrt(n*(n+1)/2))-floor(sqrt((n-1)*n/2)), n=1..200); # Robert Israel, Mar 14 2023
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MATHEMATICA
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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]
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PROG
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(PARI) a(n) = my(r); sqrtint((n^2+n)>>1, &r); r<n; \\ Kevin Ryde, May 24 2023
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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