cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A153194 Numbers such that the numerator of floor(sqrt(n))/n, when reduced to its lowest terms, is equal to 3.

Original entry on oeis.org

10, 11, 13, 14, 38, 40, 44, 46, 84, 87, 93, 96, 148, 152, 160, 164, 230, 235, 245, 250, 330, 336, 348, 354, 448, 455, 469, 476, 584, 592, 608, 616, 738, 747, 765, 774, 910, 920, 940, 950, 1100, 1111, 1133, 1144, 1308, 1320, 1344, 1356
Offset: 1

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Author

Anthony C Robin, Dec 20 2008

Keywords

Comments

Previous name was: "For each number of the sequence, n, consider the fraction of squares from 1 to n inclusive. For numbers in this sequence, that fraction, when reduced to its lowest terms will always have 3 in the numerator.".
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 3/k. Another sequence (A153192) looks at the 2/k case.

Examples

			For 38, there are 6 squares below it and 6/38=3/19.
For 164 there are 12 squares below it and 12/164=3/41.
		

Crossrefs

Cf. A153192.

Programs

  • Mathematica
    CoefficientList[Series[(10 + x + 2 x^2 + x^3 + 4 x^4)/((1 - x) (1 - x^4)^2), {x, 0, 60}], x] (* Vincenzo Librandi, Mar 29 2014 *)
  • PARI
    isok(n) = numerator(sqrtint(n)/n) == 3 \\ Michel Marcus, Aug 05 2013
    
  • PARI
    Vec(-x*(4*x^4+x^3+2*x^2+x+10)/((x-1)^3*(x+1)^2*(x^2+1)^2) + O(x^100)) \\ Colin Barker, Mar 28 2014

Formula

G.f.: -x*(4*x^4+x^3+2*x^2+x+10) / ((x-1)^3*(x+1)^2*(x^2+1)^2). - Colin Barker, Mar 28 2014
G.f.: x * (10 + x + 2*x^2 + x^3 + 4*x^4) / ((1 - x) * (1 - x^4)^2). - Michael Somos, Mar 28 2014
a(n+4) = 2*a(n) - a(n-4) + 18 if n>0. - Michael Somos, Mar 28 2014

Extensions

Edited and more terms added by Michel Marcus, Aug 05 2013