A154149 Indices k such that 22 plus the k-th triangular number is a perfect square.

2, 12, 27, 77, 162, 452, 947, 2637, 5522, 15372, 32187, 89597, 187602, 522212, 1093427, 3043677, 6372962, 17739852, 37144347, 103395437, 216493122, 602632772, 1261814387, 3512401197, 7354393202, 20471774412, 42864544827, 119318245277, 249832875762

Offset

1,1

Links

Colin Barker, Table of n, a(n) for n = 1..500

F. T. Adams-Watters, SeqFan Discussion, Oct 2009

Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Formula

{k: 22+k*(k+1)/2 in A000290}

a(n)= +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5).

G.f.: x*(-2-10*x-3*x^2+10*x^3+3*x^4)/((x-1) * (x^2-2*x-1) * (x^2+2*x-1)).

G.f.: ( 6 + (10+25*x)/(x^2-2*x-1) - 5/(x^2+2*x-1) + 1/(x-1) )/2.

Example

2*(2+1)/2+22 = 5^2. 12*(12+1)/2+22 = 10^2. 27*(27+1)/2+22 = 20^2. 77*(77+1)/2+22 = 55^2.

Mathematica

Join[{2, 12}, Select[Range[0, 10^5], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 22 &]] (* or *) LinearRecurrence[{1, 6, -6, -1, 1}, {2, 12, 27, 77, 162}, 25] (* G. C. Greubel, Sep 03 2016 *)

Prog

(PARI) Vec(x*(-2-10*x-3*x^2+10*x^3+3*x^4)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)) + O(x^30)) \\ Colin Barker, Jul 11 2015

Crossrefs

Cf. A000217, A000290, A006451.

Sequence in context: A294170 A102960 A166151 * A354780 A119201 A164876

Adjacent sequences: A154146 A154147 A154148 * A154150 A154151 A154152

Keyword

nonn,less,easy

Author

R. J. Mathar, Oct 18 2009

Extensions

Extended by D. S. McNeil, Dec 04 2010

Status

approved