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%I #19 Aug 16 2015 12:03:55
%S 1,10,493,3382,158905,1089154,51167077,350704366,16475640049,
%T 112925716858,5305104928861,36361730124070,1708227311453353,
%U 11708364174233842,550043889183050965,3770056902373173214,177112424089630957537,1213946614199987541226
%N Indices of triangular numbers which are also heptagonal.
%C From _Ant King_, Oct 19 2011: (Start)
%C lim(n->Infinity,a(2n+1)/a(2n))=1/2(47+21*sqrt(5)).
%C lim(n->Infinity,a(2n)/a(2n-1))=1/2(7+3*sqrt(5)).
%C (End)
%H Colin Barker, <a href="/A039835/b039835.txt">Table of n, a(n) for n = 1..798</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HeptagonalTriangularNumber.html">Heptagonal Triangular Number</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,322,-322,-1,1).
%F G.f.: (-2x^4-9x^3+161x^2+9x+1)/[(1-x)(1-18x+x^2)(1+18x+x^2)].
%F a(n+2) = 322*a(n+1)-a(n)+160 a(n+1) = 161*a(n)+80+36*(20*a(n)^2+20*a(n)+9)^0.5 - _Richard Choulet_, Sep 29 2007
%F From _Ant King_, Oct 19 2011: (Start)
%F a(n) = a(n-1)+322a(n-2)-322a(n-3)-a(n-4)+a(n-5).
%F a(n) = 1/20*sqrt(5)*(( sqrt(5)-(-1)^n)*(2+ sqrt(5))^(2n-1)+( sqrt(5)+(-1)^n)*(2- sqrt(5))^(2n-1)-2* sqrt(5)).
%F a(n) = floor(1/20* sqrt(5)*(sqrt(5)-(-1)^n)*(2+ sqrt(5))^(2n-1))(End)
%t LinearRecurrence[{1,322,-322,-1,1},{1,10,493,3382,158905},16] (* _Ant King_, Oct 19 2011 *)
%o (PARI) Vec((-2*x^4-9*x^3+161*x^2+9*x+1)/((1-x)*(1-18*x+x^2)*(1+18*x+x^2))+O(x^99))
%Y Cf. A046193, A046194.
%K nonn,easy
%O 1,2
%A _Eric W. Weisstein_
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