cp's OEIS Frontend

A305292 Numbers k such that k-1 is a square and k+1 is a triangular number.

%I A305292 #18 Jun 14 2018 16:45:02
%S A305292 2,5,65,170,2210,5777,75077,196250,2550410,6666725,86638865,226472402,
%T A305292 2943171002,7693394945,99981175205,261348955730,3396416785970,
%U A305292 8878171099877,115378189547777,301596468440090,3919462027838450,10245401755863185,133146330756959525
%N A305292 Numbers k such that k-1 is a square and k+1 is a triangular number.
%C A305292 It is easy to prove that there are no numbers k such that k-1 is a triangular number and k+1 is a square.
%H A305292 Colin Barker, <a href="/A305292/b305292.txt">Table of n, a(n) for n = 1..1000</a>
%H A305292 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,34,-34,-1,1).
%F A305292 G.f.: x*(2 + 3*x - 8*x^2 + 3*x^3 + 2*x^4)/((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)).
%F A305292 a(n) = a(-n-1) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5).
%F A305292 a(n) = 35*a(n-2) - 35*a(n-4) + a(n-6) = 34*a(n-2) - a(n-4) + 2.
%F A305292 a(n) = (-2 + (13*sqrt(2) + 7*(-1)^n)*(1 + sqrt(2))^(2*n+1) - (13*sqrt(2) - 7* (-1)^n)*(1 - sqrt(2))^(2*n+1))/32.
%F A305292 a(n) = A214838(n) - 1.
%F A305292 a(n) = A077241(n-1)^2 + 1.
%t A305292 LinearRecurrence[{1, 34, -34, -1, 1}, {2, 5, 65, 170, 2210}, 25]
%o A305292 (PARI) Vec(x*(2 + 3*x - 8*x^2 + 3*x^3 + 2*x^4)/((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^30)) \\ _Colin Barker_, Jun 14 2018
%Y A305292 Cf. A000045, A000290, A077241, A214838.
%K A305292 nonn,easy
%O A305292 1,1
%A A305292 _Bruno Berselli_, Jun 11 2018