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Numbers n such that the hexagonal number H(n) is equal to the sum of the pentagonal numbers P(m), P(m+1), P(m+2) and P(m+3) for some m.
2

%I #14 Sep 08 2022 08:46:10

%S 8,1480,287064,55688888,10803357160,2095795600104,406573543062968,

%T 78873171558615640,15300988708828371144,2968312936341145386248,

%U 575837408661473376560920,111709488967389493907432184,21671065022264900344665282728,4204074904830423277371157417000

%N Numbers n such that the hexagonal number H(n) is equal to the sum of the pentagonal numbers P(m), P(m+1), P(m+2) and P(m+3) for some m.

%C Also positive integers y in the solutions to 12*x^2-4*y^2+32*x+2*y+36 = 0, the corresponding values of x being A252762.

%H Colin Barker, <a href="/A252763/b252763.txt">Table of n, a(n) for n = 1..437</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (195,-195,1).

%F a(n) = 195*a(n-1)-195*a(n-2)+a(n-3).

%F G.f.: -8*x*(3*x^2-10*x+1) / ((x-1)*(x^2-194*x+1)).

%F a(n) = (6+(285-164*sqrt(3))*(97+56*sqrt(3))^n+(97+56*sqrt(3))^(-n)*(285+164*sqrt(3)))/24. - _Colin Barker_, Mar 02 2016

%F a(n) = 194*a(n-1)-a(n-2)-48. - _Vincenzo Librandi_, Mar 03 2016

%e 8 is in the sequence because H(8) = 120 = 12+22+35+51 = P(3)+P(4)+P(5)+P(6).

%t LinearRecurrence[{195, -195, 1}, {8, 1480, 287064}, 30] (* _Vincenzo Librandi_, Mar 03 2016 *)

%o (PARI) Vec(-8*x*(3*x^2-10*x+1)/((x-1)*(x^2-194*x+1)) + O(x^100))

%o (Magma) I:=[8,1480]; [n le 2 select I[n] else 194*Self(n-1)- Self(n-2)-48: n in [1..20]]; // _Vincenzo Librandi_, Mar 03 2016

%Y Cf. A000326, A000384, A252762.

%K nonn,easy

%O 1,1

%A _Colin Barker_, Dec 21 2014