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%I #18 Sep 08 2022 08:45:07
%S 1,1,8,148,5544,351982,34100352,4692680418,871465795200,
%T 210173265448681,63895600819814400,23912071579876921820,
%U 10804489706894562201600,5800208625625936700452385,3649548011303182127557017600,2660422068287264314770502524513
%N a(n) = floor(2*binomial(n+1,2)!/(binomial(n,2)!*n*(n^2+1))).
%C Consider the harmonic progression 1, 1/2, 1/3, 1/4, 1/5, ...; then a(n) = floor(reciprocal of the sum of next n terms of this harmonic progression).
%H G. C. Greubel, <a href="/A075351/b075351.txt">Table of n, a(n) for n = 1..225</a>
%e a(4) = floor(7*8*9*10/(7+8+9+10)) = floor(5040/34) = 148.
%p a:=n->floor((n*(n+1)/2)!/(n*(n-1)/2)!/(n*(n^2+1)/2)): seq(a(n),n=1..16); # _Emeric Deutsch_, Aug 04 2005
%t Table[Floor[2*Binomial[n+1, 2]!/(Binomial[n, 2]!*n*(n^2+1))], {n, 1, 25}] (* _G. C. Greubel_, Mar 07 2019 *)
%o (PARI) k=1;for(n=0,20, p=1;s=0; for(i=k,k+n,s=s+i;p=p*i); k=k+n+1; print1(floor(p/s)","))
%o (Magma) [Floor(2*Gamma((n^2+n+2)/2)/(Gamma((n^2-n+2)/2)*n*(n^2+1))): n in [1..25]]; // _G. C. Greubel_, Mar 07 2019
%o (Sage) [floor(2*factorial((n+1)*n/2)/(factorial(n*(n-1)/2)*n*(n^2+1))) for n in (1..25)] # _G. C. Greubel_, Mar 07 2019
%Y Cf. A075350.
%K nonn
%O 1,3
%A _Amarnath Murthy_, Sep 19 2002
%E More terms from _Ralf Stephan_, Mar 31 2003
%E Name edited by _Jon E. Schoenfield_, Mar 07 2019
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