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
Examples
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.
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).
Programs
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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 *) -
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
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.
Extensions
Extended by D. S. McNeil, Dec 04 2010