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%I #20 Apr 03 2019 03:03:01
%S 0,1,2,20,55,667,1856,22646,63037,769285,2141390,26133032,72744211,
%T 887753791,2471161772,30157495850,83946756025,1024467105097,
%U 2851718543066,34801724077436,96874483708207,1182234151527715,3290880727535960,40161159427864862
%N Indices of decagonal numbers (A001107) that are also triangular (A000217).
%C For n>0, a(n) = (A055979(n) - A056161(n))/2, with those two sequences related through the Diophantine equation 2x^2 + 3x + 2 = r^2. - _Richard R. Forberg_, Nov 24 2013
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1, 34, -34, -1, 1).
%F For n>5, a(n) = 34*a(n-2) - a(n-4) - 12.
%F For n>6, a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5).
%F For n>1, a(n) = 1/16 * ((2*sqrt(2) + (-1)^n)*(1 + sqrt(2))^(2*n - 3) - (2*sqrt(2) - (-1)^n)*(1 - sqrt(2))^(2*n - 3) + 6).
%F For n>1, a(n) = ceiling (1/16*(2*sqrt(2) + (-1)^n)*(1 + sqrt(2))^(2*n - 3)).
%F G.f.: ( 1 - 33*x^2 + 18*x^3 + 2*x^4 ) / ((1 - x ) * (1 - 6*x + x^2 ) * (1 + 6*x + x^2)).
%F lim (n -> Infinity, a(2n+1)/a(2n)) = 1/7*(43 + 30*sqrt(2)).
%F lim (n -> Infinity, a(2n)/a(2n-1)) = 1/7*(11 + 6*sqrt(2)).
%e The third number which is both decagonal (A001107) and triangular (A000217) is A133216(3)=10. As this is the second decagonal number, we have a(3) = 2.
%t LinearRecurrence[{1, 34, -34, -1, 1} , {0, 1, 2, 20, 55, 667}, 24] (* first term 0 corrected by _Georg Fischer_, Apr 02 2019 *)
%Y Cf. A000217, A001107, A077443, A077442, A133216, A133218.
%K nonn
%O 1,3
%A _Richard Choulet_, Oct 11 2007; _Ant King_, Nov 04 2011
%E Entry revised by _Max Alekseyev_, Nov 06 2011
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