%I #13 Nov 12 2017 17:19:09
%S 1,6,46,361,2841,22366,176086,1386321,10914481,85929526,676521726,
%T 5326244281,41933432521,330141215886,2599196294566,20463429140641,
%U 161108236830561,1268402465503846,9986111487200206,78620489432097801,618977803969582201,4873201942324559806
%N Indices of pentagonal numbers (A000326) which are also centered pentagonal numbers (A005891).
%C Also positive integers x in the solutions to 3*x^2-5*y^2-x+5*y-2 = 0, the corresponding values of y being A253470.
%H Colin Barker, <a href="/A253654/b253654.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (9,-9,1).
%F a(n) = 9*a(n-1)-9*a(n-2)+a(n-3).
%F G.f.: -x*(x^2-3*x+1) / ((x-1)*(x^2-8*x+1)).
%F a(n) = (2-(-5+sqrt(15))*(4+sqrt(15))^n+(4-sqrt(15))^n*(5+sqrt(15)))/12. - _Colin Barker_, Mar 03 2016
%e 6 is in the sequence because the 6th pentagonal number is 51, which is also the 5th centered pentagonal number.
%t LinearRecurrence[{9,-9,1},{1,6,46},30] (* _Harvey P. Dale_, Nov 12 2017 *)
%o (PARI) Vec(-x*(x^2-3*x+1)/((x-1)*(x^2-8*x+1)) + O(x^100))
%Y Cf. A000326, A005891, A128917, A253470.
%K nonn,easy
%O 1,2
%A _Colin Barker_, Jan 07 2015