A154151 Indices k such that 25 plus the k-th triangular number is a perfect square.

0, 18, 21, 111, 128, 650, 749, 3791, 4368, 22098, 25461, 128799, 148400, 750698, 864941, 4375391, 5041248, 25501650, 29382549, 148634511, 171254048, 866305418, 998141741, 5049197999, 5817596400, 29428882578, 33907436661, 171524097471, 197627023568

Offset

1,2

Links

Harvey P. Dale, 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

{k: 25+k*(k+1)/2 in A000290}.

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*(-18-3*x+18*x^2+x^3)/( (x-1) * (x^2+2*x-1) * (x^2-2*x-1)).

G.f.: ( 2 + 1/(x-1) + (10+29*x)/(x^2-2*x-1) + (-9+8*x)/(x^2+2*x-1) )/2. (End)

The first conjecture is true for the first 1000 terms of the sequence. - Harvey P. Dale, Jun 15 2013

Example

0*(0+1)/2+25 = 5^2. 18*(18+1)/2+25 = 14^2. 21*(21+1)/2+25 = 16^2. 111*(111+1)/2+25 = 79^2.

Mathematica

Join[{0}, Select[Range[0, 10^5], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 25 &]] (* or *) LinearRecurrence[{1, 6, -6, -1, 1}, {0, 18, 21, 111, 128}, 25] (* G. C. Greubel_, Sep 03 2016 *)

Prog

(PARI) for(n=1, 10^10, if(issquare(25+n*(n+1)/2), print1(n, ", ")))

Crossrefs

Cf. A000217, A000290, A006451.

Sequence in context: A303298 A358592 A121851 * A296008 A049734 A096282

Adjacent sequences: A154148 A154149 A154150 * A154152 A154153 A154154

Keyword

nonn,less,easy

Author

R. J. Mathar, Oct 18 2009

Status

approved