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A256173
Numbers k such that ceiling(sqrt(k))^2 - k is a square.
1
0, 1, 3, 4, 5, 8, 9, 12, 15, 16, 21, 24, 25, 27, 32, 35, 36, 40, 45, 48, 49, 55, 60, 63, 64, 65, 72, 77, 80, 81, 84, 91, 96, 99, 100, 105, 112, 117, 120, 121, 128, 135, 140, 143, 144, 153, 160, 165, 168, 169, 171, 180, 187, 192, 195, 196, 200, 209, 216, 221, 224, 225, 231, 240, 247, 252, 255, 256, 264, 273, 280, 285, 288, 289, 299
OFFSET
1,3
COMMENTS
Numbers k such that A068527(k) is a square. k is in the sequence if and only if k - ceiling(sqrt(k))^2 + ceiling(sqrt(ceiling(sqrt(k))^2 - k))^2 = 0.
A000290 is a subsequence since for a square k, ceiling(sqrt(k))^2 - k = 0, a square too.
Also, numbers k such that A249298(k) is 1.
Also, numbers k such that A249142(k) is 0.
The only prime numbers in the sequence are 3 and 5.
No number from A016825 appears in the sequence.
If p and q are terms of A065091 and if q satisfies the inequality p - 2*sqrt(2p) + 2 < q < p + 2*sqrt(2p) + 2, then p*q is in the sequence. Thus infinitely many numbers from A046315 appear in the sequence.
EXAMPLE
Ceiling(sqrt(27))^2 - 27 = 9 = 3^2, so 27 is in the sequence.
MATHEMATICA
Flatten[Position[Table[n - Ceiling[Sqrt[n]]^2 + Ceiling[Sqrt[-n + Ceiling[Sqrt[n]]^2]]^2, {n, 0, 300}], 0]] - 1
Select[Range[0, 300], IntegerQ[Sqrt[Ceiling[Sqrt[#]]^2-#]]&] (* Harvey P. Dale, Sep 06 2023 *)
PROG
(PARI) isok(n) = issquare(ceil(sqrt(n))^2-n); \\ Michel Marcus, Mar 18 2015
(Magma) [n: n in [0..200] | IsSquare(Ceiling(Sqrt(n))^2-n)]; // Vincenzo Librandi, Mar 18 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Valtteri Raiko, Mar 17 2015
STATUS
approved