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A184809
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a(n) = n + floor(sqrt(3/2)*n).
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5
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2, 4, 6, 8, 11, 13, 15, 17, 20, 22, 24, 26, 28, 31, 33, 35, 37, 40, 42, 44, 46, 48, 51, 53, 55, 57, 60, 62, 64, 66, 68, 71, 73, 75, 77, 80, 82, 84, 86, 88, 91, 93, 95, 97, 100, 102, 104, 106, 109, 111, 113, 115, 117, 120, 122, 124, 126, 129, 131, 133, 135
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OFFSET
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1,1
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COMMENTS
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This is the Beatty sequence for 1 + sqrt(3/2).
Also, a(n) is the position of 3*n^2 in the sequence obtained by arranging all the numbers in the sets {2*h^2, h >= 1} and {3*k^2, k >= 1} in increasing order. - Clark Kimberling, Oct 20 2014
Also, numbers n such that floor((n+1)*sqrt(6)) - floor(n*sqrt(6)) = 3. - Clark Kimberling, Jul 15 2015
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LINKS
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FORMULA
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a(n) = n + floor(r*n), where r = sqrt(3/2).
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MATHEMATICA
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PROG
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(PARI) main(size)={return(vector(size, n, n+floor(sqrt(3/2)*n)))}/* Anders Hellström, Jul 15 2015 */
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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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