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A232091
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Smallest square or promic (oblong) number greater than or equal to n.
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3
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0, 1, 2, 4, 4, 6, 6, 9, 9, 9, 12, 12, 12, 16, 16, 16, 16, 20, 20, 20, 20, 25, 25, 25, 25, 25, 30, 30, 30, 30, 30, 36, 36, 36, 36, 36, 36, 42, 42, 42, 42, 42, 42, 49, 49, 49, 49, 49, 49, 49, 56, 56, 56, 56, 56, 56, 56, 64, 64, 64, 64, 64, 64, 64, 64, 72, 72, 72, 72, 72, 72, 72, 72, 81
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OFFSET
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0,3
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COMMENTS
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Result attributed to the students Daring, et al., in the links section.
a(n) appears in floor(sqrt(a(n)) = A000194(n) successive terms.
Counting successive equal terms give sequence: 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ... (see A008619). - Michel Marcus, Jan 10 2014
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LINKS
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FORMULA
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a(n) = ceiling(n/ceiling(sqrt(n)))*ceiling(sqrt(n)).
a(n) = min(k : k >= n, k in A002620).
a(k^2) = k^2; a(k*(k+1)) = k*(k+1).
It appears that a(n) = A216607(n) + n. (Verified for all n<10^9 by Lars Blomberg, Jan 09 2014.) This conjecture now follows from a proof given by David Applegate, Jan 10 2014 (see [Applegate]).
Sum_{n>=1} 1/a(n)^2 = 2 - Pi^2/6 + zeta(3). - Amiram Eldar, Aug 16 2022
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MATHEMATICA
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Join[{0}, Table[Ceiling[n/Ceiling[Sqrt[n]]] Ceiling[Sqrt[n]], {n, 100}]] (* Alonso del Arte, Nov 18 2013 *)
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PROG
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(Magma) [(Ceiling(n /Ceiling(Sqrt(n)))*Ceiling(Sqrt(n))): n in [1..80]]; // Vincenzo Librandi, Jun 22 2015
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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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EXTENSIONS
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STATUS
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approved
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