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%I #23 Jul 09 2021 17:11:51
%S 1,54,3337,206830,12820113,794640166,49254870169,3053007310302,
%T 189237198368545,11729653291539478,727049266877079081,
%U 45065324893087363534,2793323094104539460017,173140966509588359157510,10731946600500373728305593,665207548264513582795789246
%N Indices of pentagonal numbers that are also heptagonal.
%C As n increases, this sequence is approximately geometric with common ratio r = lim_{n->infinity} a(n)/a(n-1) = (4+sqrt(15))^2 = 31 + 8*sqrt(15). - _Ant King_, Dec 15 2011
%H Colin Barker, <a href="/A046199/b046199.txt">Table of n, a(n) for n = 1..558</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HeptagonalPentagonalNumber.html">Heptagonal Pentagonal Number.</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (63,-63,1).
%F From _Ant King_, Dec 15 2011: (Start)
%F a(n) = 63*a(n-1) - 63*a(n-2) + a(n-3).
%F a(n) = 62*a(n-1) - a(n-2) - 10.
%F a(n) = (1/60)*((3*sqrt(15)-5)*(4+sqrt(15))^(2*n-1) - (3*sqrt(15)+5)*(4-sqrt(15))^(2*n-1)+10).
%F a(n) = ceiling((1/60)*(3*sqrt(15)-5)*(4+sqrt(15))^(2*n-1)).
%F G.f.: x*(1-9*x-2*x^2)/((1-x)*(1-62*x+x^2)).
%F (End)
%t LinearRecurrence[{63, -63, 1}, {1, 54, 3337}, 14] (* _Ant King_, Dec 15 2011 *)
%t CoefficientList[Series[x (1-9x-2x^2)/((1-x)(1-62x+x^2)),{x,0,20}],x] (* _Harvey P. Dale_, Jul 09 2021 *)
%o (PARI) Vec(x*(2*x^2+9*x-1)/((x-1)*(x^2-62*x+1)) + O(x^30)) \\ _Colin Barker_, Jun 23 2015
%Y Cf. A046198, A048900.
%K nonn,easy
%O 1,2
%A _Eric W. Weisstein_
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