login
Numbers n such that the sum of the octagonal numbers N(n) and N(n+1) is equal to another octagonal number.
2

%I #6 Jun 13 2015 00:55:19

%S 0,30,368,34814,424864,40175518,490292880,46362513150,565797558848,

%T 53502299999774,652929892617904,61741607837226238,753480530283502560,

%U 71249761941859079070,869515879017269336528,82222163539297540020734,1003420570905398530850944

%N Numbers n such that the sum of the octagonal numbers N(n) and N(n+1) is equal to another octagonal number.

%C Also nonnegative integers x in the solutions to 12*x^2-6*y^2+4*x+4*y+2 = 0, the corresponding values of y being A251896.

%H Colin Barker, <a href="/A251895/b251895.txt">Table of n, a(n) for n = 1..653</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,1154,-1154,-1,1).

%F a(n) = a(n-1)+1154*a(n-2)-1154*a(n-3)-a(n-4)+a(n-5).

%F G.f.: 2*x^2*(x^3+87*x^2-169*x-15) / ((x-1)*(x^2-34*x+1)*(x^2+34*x+1)).

%e 30 is in the sequence because N(30)+N(31) = 2640+2821 = 5461 = N(43).

%o (PARI) concat(0, Vec(2*x^2*(x^3+87*x^2-169*x-15)/((x-1)*(x^2-34*x+1)*(x^2+34*x+1)) + O(x^100)))

%Y Cf. A000567, A251896.

%K nonn,easy

%O 1,2

%A _Colin Barker_, Dec 10 2014