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A153192
Numbers such that the numerator of floor(sqrt(n))/n, when reduced to its lowest terms, is equal to 2.
2
5, 7, 18, 22, 39, 45, 68, 76, 105, 115, 150, 162, 203, 217, 264, 280, 333, 351, 410, 430, 495, 517, 588, 612, 689, 715, 798, 826, 915, 945, 1040, 1072, 1173, 1207, 1314, 1350, 1463, 1501, 1620, 1660, 1785, 1827, 1958
OFFSET
1,1
COMMENTS
Previous name was: "For each term in this sequence, n say, consider the fraction of square numbers between 1 & n, inclusive, when reduced to its lowest terms. This fraction will always have a numerator of 2 for numbers in this sequence".
To obtain similar fractions as above with a numerator of 1, for example 1/5 are square, there are three possible numbers, namely 15, 20, 25. In general it is fairly easy to show that for 1/k of the numbers up to n (1 to n inclusive) to be square, n takes one of the three values, k(k-2), k(k-1), k^2. This sequence looks at obtaining fractions of the form 2/k. Another sequence (A153194) looks at the 3/k case.
Alternately, numbers of the form 4n^2+n or 4n^2+3n. - Charles R Greathouse IV, Aug 05 2013
FORMULA
From Colin Barker, Mar 28 2014: (Start)
a(n) = (2*n+3)*(2*n-(-1)^n+1)/4.
a(n) = a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4)+a(n-5).
G.f.: -x*(x^2+2*x+5) / ((x-1)^3*(x+1)^2). (End).
EXAMPLE
22 has 4 square numbers below it and 4/22=2/11.
76 has 8 square numbers below it and 8/76=2/19.
MATHEMATICA
LinearRecurrence[{1, 2, -2, -1, 1}, {5, 7, 18, 22, 39}, 50] (* Harvey P. Dale, Sep 23 2022 *)
PROG
(PARI) isok(n) = numerator(sqrtint(n)/n) == 2 \\ Michel Marcus, Aug 05 2013
(PARI) for(n=1, 9, print1(4*n^2+n", "4*n^2+3*n", ")) \\ Charles R Greathouse IV, Aug 05 2013
CROSSREFS
Cf. A153194.
Sequence in context: A138919 A345909 A214414 * A370775 A263878 A297937
KEYWORD
nonn,easy
AUTHOR
Anthony C Robin, Dec 20 2008
EXTENSIONS
Definition simplified and more terms added by Michel Marcus, Aug 05 2013
STATUS
approved