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