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%I #19 Sep 08 2022 08:46:17
%S 1,6,12,37,71,216,414,1259,2413,7338,14064,42769,81971,249276,477762,
%T 1452887,2784601,8468046,16229844,49355389,94594463,287664288,
%U 551336934,1676630339,3213427141,9772117746,18729225912,56956076137,109161928331,331964339076
%N Values of k such that k^2 + 9 is a triangular number (A000217).
%H Colin Barker, <a href="/A276601/b276601.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,6,0,-1).
%F a(n) = 6*a(n-2) - a(n-4) for n>4.
%F G.f.: x*(1+x)*(1+5*x+x^2) / ((1+2*x-x^2)*(1-2*x-x^2)).
%F a(n) = (1/4)*(5*P(n+1) - P(n) + (-1)^n*(P(n-1) + 5*P(n-2))), where P(n) = A000129(n). - _G. C. Greubel_, Sep 15 2021
%e 6 is in the sequence because 6^2+9 = 45, which is a triangular number.
%t CoefficientList[Series[(1+x)*(1+5*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)), {x, 0, 30}], x] (* _Wesley Ivan Hurt_, Sep 07 2016 *)
%t LinearRecurrence[{0,6,0,-1}, {1,6,12,37}, 31] (* _G. C. Greubel_, Sep 15 2021 *)
%o (PARI) Vec(x*(1+x)*(1+5*x+x^2) / ((1+2*x-x^2)*(1-2*x-x^2)) + O(x^40))
%o (Magma) I:=[1,6,12,37]; [n le 2 select I[n] else 6*Self(n-2) - Self(n-4): n in [1..31]]; // _G. C. Greubel_, Sep 15 2021
%o (Sage)
%o def P(n): return lucas_number1(n, 2, -1)
%o [(1/4)*(5*P(n+1) - P(n) + (-1)^n*(P(n-1) + 5*P(n-2))) for n in (1..30)] # _G. C. Greubel_, Sep 15 2021
%Y Cf. A000129, A000217, A230044.
%Y Cf. A001109 (k=0), A106328 (k=1), A077241 (k=2), A276598 (k=3), A276599 (k=5), A276600 (k=6), A276602 (k=10), where k is the value added to n^2.
%K nonn,easy
%O 1,2
%A _Colin Barker_, Sep 07 2016