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%I #10 Nov 13 2017 10:43:22
%S 3,234,1617,112948,779551,54440862,375742125,26240382696,181106924859,
%T 12647810018770,87293162040073,6096218188664604,42075122996390487,
%U 2938364519126320518,20280121991098174821,1416285602000697825232,9774976724586323873395
%N Numbers n such that the sum of the pentagonal numbers P(n) and P(n+1) is equal to the heptagonal number H(m) for some m.
%C Also positive integers x in the solutions to 6*x^2-5*y^2+4*x+3*y+2 = 0, the corresponding values of y being A252586.
%H Colin Barker, <a href="/A252585/b252585.txt">Table of n, a(n) for n = 1..745</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,482,-482,-1,1).
%F a(n) = a(n-1)+482*a(n-2)-482*a(n-3)-a(n-4)+a(n-5).
%F G.f.: x*(11*x^3+63*x^2-231*x-3) / ((x-1)*(x^2-22*x+1)*(x^2+22*x+1)).
%e 3 is in the sequence because P(3)+P(4) = 12+22 = 34 = H(4).
%t LinearRecurrence[{1, 482, -482, -1, 1}, {3, 234, 1617, 112948, 779551}, 20] (* _Jean-François Alcover_, Nov 13 2017 *)
%o (PARI) Vec(x*(11*x^3+63*x^2-231*x-3)/((x-1)*(x^2-22*x+1)*(x^2+22*x+1)) + O(x^100))
%Y Cf. A000326, A000566, A252586.
%K nonn,easy
%O 1,1
%A _Colin Barker_, Dec 18 2014
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