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

%I #13 Sep 08 2022 08:46:10

%S 3,853,165735,32151993,6237321163,1210008153885,234735344532783,

%T 45537446831206273,8834029949909484435,1713756272835608774373,

%U 332459882900158192744183,64495503526357853783597385,12511795224230523475825148763,2427223777997195196456295262893

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

%C Also positive integers x in the solutions to 12*x^2-4*y^2+32*x+2*y+36 = 0, the corresponding values of y being A252763.

%H Colin Barker, <a href="/A252762/b252762.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.: x*(15*x^2-268*x-3) / ((x-1)*(x^2-194*x+1)).

%F a(n) = -4/3+1/24*(97+56*sqrt(3))^(-n)*(-164-95*sqrt(3)+(97+56*sqrt(3))^(2*n)*(-164+95*sqrt(3))). - _Colin Barker_, Mar 02 2016

%F a(n) = 194*a(n-1)-a(n-2)+256. - _Vincenzo Librandi_, Mar 03 2016

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

%t LinearRecurrence[{195, -195, 1}, {3, 853, 165735}, 30] (* _Vincenzo Librandi_, Mar 03 2016 *)

%o (PARI) Vec(x*(15*x^2-268*x-3)/((x-1)*(x^2-194*x+1)) + O(x^100))

%o (Magma) I:=[3,853]; [n le 2 select I[n] else 194*Self(n-1) - Self(n-2)+256: n in [1..20]]; // _Vincenzo Librandi_, Mar 03 2016

%Y Cf. A000326, A000384, A252763.

%K nonn,easy

%O 1,1

%A _Colin Barker_, Dec 21 2014