%I #14 Mar 28 2024 09:01:59
%S 1,540,2887450,1123674201,6004054625647,2336525434757970,
%T 12484603034492528512,4858482201068079159687,
%U 25960009135002449017962445,10102543266574986692211140472,53980256514964477791853933850326,21006844571867038996088473395797925
%N Numbers which are both heptagonal and decagonal.
%C As n increases, the ratios of consecutive terms settle into an approximate 2-cycle with a(n)/a(n-1) bounded above and below by 1/81*(216401+68432*sqrt(10)) and 1/81*(15761+4984*sqrt(10)) respectively.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2079362,-2079362,-1,1).
%F G.f.: x(1+539*x+807548*x^2+10633*x^3+27*x^4) / ((1-x)*(1-1442*x+x^2)*(1+1442*x+x^2)).
%F a(n) = 2079362*a(n-2)-a(n-4)+818748.
%F a(n) = a(n-1)+2079362*a(n-2)-2079362*a(n-3)-a(n-4)+a(n-5).
%F a(n) = 1/320*((11-2*sqrt(10)*(-1)^n)*(1+sqrt(10))* (3+sqrt(10))^(4*n-3)+(11+2*sqrt(10)*(-1)^n)*(1-sqrt(10))*(3-sqrt(10))^(4*n-3)-126).
%F a(n) = floor(1/320*(11-2*sqrt(10)*(-1)^n)*(1+sqrt(10))* (3+sqrt(10))^(4*n-3)).
%e The second number that is both heptagonal and decagonal is 540. Hence a(2)=540.
%t LinearRecurrence[{1, 2079362, -2079362, -1, 1}, {1, 540, 2887450, 1123674201, 6004054625647}, 15]
%Y Cf. A203409, A203410, A001107, A000566.
%K nonn,easy
%O 1,2
%A _Ant King_, Jan 02 2012