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Numbers n such that the sum of the octagonal numbers N(n) and N(n+1) is equal to the sum of the pentagonal numbers P(m) and P(m+1) for some m.
2

%I #12 Sep 14 2020 11:41:42

%S 232,267920,309179640,356793036832,411738855324680,475146282251644080,

%T 548318397979541943832,632758956122109151538240,

%U 730203287046515981333185320,842653960492723320349344321232,972421940205315665167162013516600,1122174076342973784879584614253835360

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

%C Also nonnegative integers x in the solutions to 6*x^2-3*y^2+2*x-2*y = 0, the corresponding values of y being A252004.

%H Colin Barker, <a href="/A252003/b252003.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*(5*x-29) / ((x-1)*(x^2-1154*x+1)).

%F a(n) = (-2 - (sqrt(2)-1)*(577+408*sqrt(2))^(-n) + (sqrt(2)+1)*(577+408*sqrt(2))^n) / 12. - _Colin Barker_, May 30 2017

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

%t LinearRecurrence[{1155,-1155,1},{232,267920,309179640},20] (* _Harvey P. Dale_, Sep 14 2020 *)

%o (PARI) Vec(8*x*(5*x-29)/((x-1)*(x^2-1154*x+1)) + O(x^100))

%Y Cf. A000326, A000567, A252004.

%K nonn,easy

%O 1,1

%A _Colin Barker_, Dec 12 2014