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A324178
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Integers k such that floor(sqrt(k)) + floor(sqrt(k/5)) divides k.
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2
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1, 2, 3, 4, 6, 12, 24, 28, 35, 40, 45, 50, 60, 66, 77, 91, 112, 128, 153, 190, 200, 220, 231, 276, 312, 338, 378, 406, 435, 450, 480, 496, 512, 561, 578, 648, 703, 722, 741, 780, 800, 840, 882, 903, 946, 968, 990, 1058, 1152, 1176, 1250, 1300, 1352, 1378
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OFFSET
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1,2
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COMMENTS
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This sequence is infinite for the same reason that A324175 is: if x > y satisfies x^2 - 5*y^2 = -1 (x=A075796(j), y=A007805(j-1), j>0), then x < 5*y. Let k = 5*y^2 + m. By the pigeonhole principle there exists a number m belonging to [0, 2*x - 1] such that x + y | 5*y^2 + m, so such a k is a term.
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LINKS
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MATHEMATICA
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Select[Range[1378], Mod[#, Floor@ Sqrt@ # + Floor@ Sqrt[#/5]] == 0 &] (* Giovanni Resta, Apr 05 2019 *)
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PROG
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(PARI) is(n) = n%(floor(sqrt(n)) + floor(sqrt(n/5))) == 0;
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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