OFFSET
1,2
COMMENTS
Also numbers n such that (ceiling(sqrt(n*(n+1)/2)))^2 - n*(n+1)/2 = 4. - Ctibor O. Zizka, Nov 10 2009
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
F. T. Adams-Watters, SeqFan Discussion, Oct 2009
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).
FORMULA
a(n) = a(n-1) + 6*a(n-2) - 6*a(n-3) - a(n-4) + a(n-5).
G.f.: x^2*(6 +3*x -6*x^2 -x^3)/((1-x)*(x^2-2*x-1)*(x^2+2*x-1)) = 1 + 1/2*(4+11*x)/(x^2-2*x-1) + 1/2/(x-1) + 1/2*(-3+2*x)/(x^2+2*x-1).
For n>4, a(n) = 6*a(n-2) - a(n-4) + 2. - Ctibor O. Zizka, Nov 10 2009
EXAMPLE
0*(0+1)/2+4 = 2^2. 6*(6+1)/2+4 = 5^2. 9*(9+1)/2+4 = 7^2. 39*(39+1)/2+4 = 28^2.
MAPLE
a := proc (n) if type(sqrt(4+(1/2)*n*(n+1)), integer) = true then n else end if end proc: seq(a(n), n = 0 .. 10^7); # Emeric Deutsch, Oct 31 2009
MATHEMATICA
LinearRecurrence[{1, 6, -6, -1, 1}, {0, 6, 9, 39, 56}, 40] (* Vincenzo Librandi, Dec 11 2012 *)
Join[{0}, Select[Range[0, 1000], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 4 &] ] (* G. C. Greubel, Sep 03 2016 *)
Join[{0}, Position[Accumulate[Range[66000]]+4, _?(IntegerQ[Sqrt[#]]&)]//Flatten] (* The program generates the first 13 terms of the sequence. *) (* Harvey P. Dale, Feb 18 2023 *)
PROG
(Magma) I:=[0, 6, 9, 39, 56]; [n le 5 select I[n] else Self(n-1)+6*Self(n-2)-6*Self(n-3)-Self(n-4)+Self(n-5): n in [1..40]]; // Vincenzo Librandi, Dec 11 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
R. J. Mathar, Oct 18 2009
EXTENSIONS
a(17)-a(18) from Emeric Deutsch, Oct 31 2009
a(19)-a(25) from Donovan Johnson, Nov 01 2010
More terms from Max Alekseyev, Jan 24 2012
STATUS
approved