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%I #25 Sep 01 2022 06:32:34
%S 125,500,1250,2500,4375,7000,10500,15000,20625,27500,35750,45500,
%T 56875,70000,85000,102000,121125,142500,166250,192500,221375,253000,
%U 287500,325000,365625,409500,456750,507500,561875,620000,682000,748000,818125,892500,971250,1054500
%N a(n) = binomial(n+2,3)*5^3.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F G.f.: 125*x / (1-x)^4.
%F a(n) = C(n+2,3)*5^3, n>=1.
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(1)=125, a(2)=500, a(3)=1250, a(4)=2500. - _Harvey P. Dale_, Oct 20 2012
%F From _Amiram Eldar_, Sep 01 2022: (Start)
%F Sum_{n>=1} 1/a(n) = 3/250.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = 12*log(2)/125 - 3/50. (End)
%p seq(binomial(n+2,3)*5^3, n=1..44);
%t With[{c=5^3},c*Binomial[Range[40]+2,3]] (* _Harvey P. Dale_, Oct 20 2012 *)
%t LinearRecurrence[ {4,-6,4,-1},{125,500,1250,2500},40] (* _Harvey P. Dale_, Oct 20 2012 *)
%K nonn,easy
%O 1,1
%A _Zerinvary Lajos_, Aug 09 2008
%E Offset corrected by _Harvey P. Dale_, Oct 20 2012
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