OFFSET
0,3
COMMENTS
Conjecture: The sequence lists all nonnegative m, in increasing order, such that floor(m/2)*floor(m/3) is a square.
This conjecture has been proved by Robert Israel (see paper in Links section).
LINKS
Robert Israel, Table of n, a(n) for n = 0..1000
Robert Israel, Proof of a conjecture.
Index entries for linear recurrences with constant coefficients, signature (0,99,0,-99,0,1).
FORMULA
G.f.: x*(1 + 2*x - 96*x^2 - 148*x^3 + 45*x^4 + 50*x^5 + 2*x^6)/((1 - x)*(1 + x)*(1 - 10*x + x^2)*(1 + 10*x + x^2)). (For e.g.f see Israel's paper.)
a(n) = 99*a(n-2) - 99*a(n-4) + a(n-6) for n>7.
a(n) = -a(n-1) + 98*a(n-2) + 98*a(n-3) - a(n-4) - a(n-5) - 144 for n>6.
MAPLE
gf:= x*(1 + 2*x - 96*x^2 - 148*x^3 + 45*x^4 + 50*x^5 + 2*x^6)/(1 - 99*x^2 + 99*x^4 - x^6):
S:= series(gf, x, 51):
seq(coeff(S, x, j), j=0..50); # Robert Israel, Feb 11 2016
MATHEMATICA
CoefficientList[x*(1 + 2*x - 96*x^2 - 148*x^3 + 45*x^4 + 50*x^5 + 2*x^6)/(1 - 99*x^2 + 99*x^4 - x^6) + O[x]^30, x] (* Jean-François Alcover, Feb 12 2016 *)
PROG
(SageMath) gf = x*(1 + 2*x - 96*x^2 - 148*x^3 + 45*x^4 + 50*x^5 + 2*x^6)/(1 - 99*x^2 + 99*x^4 - x^6); taylor(gf, x, 0, 30).list()
(PARI) concat(0, Vec((1 + 2*x - 96*x^2 - 148*x^3 + 45*x^4 + 50*x^5 + 2*x^6)/(1 - 99*x^2 + 99*x^4 - x^6) + O(x^30)))
(Maxima) makelist(coeff(taylor(x*(1 + 2*x - 96*x^2 - 148*x^3 + 45*x^4 + 50*x^5 + 2*x^6)/(1 - 99*x^2 + 99*x^4 - x^6), x, 0, n), x, n), n, 0, 30);
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!((1 + 2*x - 96*x^2 - 148*x^3 + 45*x^4 + 50*x^5 + 2*x^6)/(1 - 99*x^2 + 99*x^4 - x^6)));
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruno Berselli, Jan 29 2016
STATUS
approved