login
A154140
Indices k such that 6 plus the k-th triangular number is a perfect square.
4
2, 4, 19, 29, 114, 172, 667, 1005, 3890, 5860, 22675, 34157, 132162, 199084, 770299, 1160349, 4489634, 6763012, 26167507, 39417725, 152515410, 229743340, 888924955, 1339042317, 5181034322, 7804510564, 30197280979, 45488021069, 176002651554, 265123615852
OFFSET
1,1
COMMENTS
In general, indices k such that A001109(2j) plus the k-th triangular number is a perfect square may be found as follows:
b(2n-1) = A001652(n+j-1) - A001653(n-j);
b(2n) = A001652(n-j-1) + A001653(n+j);
Indices k such that A001109(2j-1) plus the k-th triangular number is a perfect square may be found as follows:
b(2n-1) = A001652(n+j-1) - A001653(n-j+1);
b(2n) = A001652(n-j) + A001653(n+j). - Charlie Marion, Mar 10 2011
FORMULA
{k: 6+k*(k+1)/2 in A000290}.
a(2*n-1) = A001652(n) - A001653(n-1).
a(2*n) = A001652(n-2) + A001653(n+1).
Conjectures: (Start)
a(n) = +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5).
G.f.: x*(2 +2*x +3*x^2 -2*x^3 -3*x^4)/((1-x)* (x^2-2*x-1)* (x^2+2*x-1))
G.f.: ( 6 + (-1 -4*x)/(x^2+2*x-1) + (6 +13*x)/(x^2-2*x-1) + 1/(x-1) )/2. (End)
a(1..4) = (2,4,19,29); a(n) = 6*a(n-2) - a(n-4) + 2, for n > 4. - Ctibor O. Zizka, Nov 10 2009
EXAMPLE
2*(2+1)/2+6 = 3^2. 4*(4+1)/2+6 = 4^2. 19*(19+1)/2+6 = 14^2. 29*(29+1)/2+6 = 21^2.
MATHEMATICA
LinearRecurrence[{1, 6, -6, -1, 1}, {2, 4, 19, 29, 114}, 40] (* Following first conjecture *) (* Harvey P. Dale, Apr 11 2016 *)
Join[{2}, Select[Range[1, 1010], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 6 &] ] (* G. C. Greubel, Sep 03 2016 *)
PROG
(Magma) [2] cat [n: n in [0..2*10^7] | (Ceiling(Sqrt(n*(n + 1)/2)) )^2 - n*(n + 1)/2 eq 6]; // Vincenzo Librandi, Sep 03 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
R. J. Mathar, Oct 18 2009
EXTENSIONS
a(17)-a(24) from Donovan Johnson, Nov 01 2010
a(25)-a(30) from Lars Blomberg, Jul 07 2015
STATUS
approved