OFFSET
1,1
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.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,2,-2,0,0,-1,1).
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
EXAMPLE
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.
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Anthony C Robin, Dec 20 2008
EXTENSIONS
Edited and more terms added by Michel Marcus, Aug 05 2013
STATUS
approved