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Numbers n such that the sum of the pentagonal numbers P(n) and P(n+1) is equal to the sum of the octagonal numbers N(m) and N(m+1) for some m.
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%I #12 May 30 2017 08:46:23

%S 328,378896,437246040,504581551648,582286673356136,671958316471429680,

%T 775439314921356494968,894856297460928923763776,

%U 1032663391830597056666902920,1191692659316211542464682206288,1375212296187516289407186599153816

%N Numbers n such that the sum of the pentagonal numbers P(n) and P(n+1) is equal to the sum of the octagonal numbers N(m) and N(m+1) for some m.

%C Also nonnegative integers y in the solutions to 6*x^2-3*y^2+2*x-2*y = 0, the corresponding values of x being A252003.

%H Colin Barker, <a href="/A252004/b252004.txt">Table of n, a(n) for n = 1..326</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1155,-1155,1).

%F a(n) = 1155*a(n-1)-1155*a(n-2)+a(n-3).

%F G.f.: -8*x*(7*x+41) / ((x-1)*(x^2-1154*x+1)).

%F a(n) = (-4 - (sqrt(2)-2)*(577+408*sqrt(2))^(-n) + (sqrt(2)+2)*(577+408*sqrt(2))^n) / 12. - _Colin Barker_, May 30 2017

%e 328 is in the sequence because P(328)+P(329) = 161212+162197 = 323409 = 161008+162401 = N(232)+N(233).

%t LinearRecurrence[{1155,-1155,1},{328,378896,437246040},30] (* _Harvey P. Dale_, Feb 21 2017 *)

%o (PARI) Vec(-8*x*(7*x+41)/((x-1)*(x^2-1154*x+1)) + O(x^100))

%Y Cf. A000326, A000567, A252003.

%K nonn,easy

%O 1,1

%A _Colin Barker_, Dec 12 2014