A075351 a(n) = floor(2*binomial(n+1,2)!/(binomial(n,2)!*n*(n^2+1))).

1, 1, 8, 148, 5544, 351982, 34100352, 4692680418, 871465795200, 210173265448681, 63895600819814400, 23912071579876921820, 10804489706894562201600, 5800208625625936700452385, 3649548011303182127557017600, 2660422068287264314770502524513

Offset

1,3

Comments

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).

Links

G. C. Greubel, Table of n, a(n) for n = 1..225

Example

a(4) = floor(7*8*9*10/(7+8+9+10)) = floor(5040/34) = 148.

Maple

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

Mathematica

Table[Floor[2*Binomial[n+1, 2]!/(Binomial[n, 2]!*n*(n^2+1))], {n, 1, 25}] (* G. C. Greubel, Mar 07 2019 *)

Prog

(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)", "))
(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
(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

Crossrefs

Cf. A075350.

Sequence in context: A116876 A218305 A244028 * A302063 A220559 A264642

Adjacent sequences: A075348 A075349 A075350 * A075352 A075353 A075354

Keyword

nonn

Author

Amarnath Murthy, Sep 19 2002

Extensions

More terms from Ralf Stephan, Mar 31 2003

Name edited by Jon E. Schoenfield, Mar 07 2019

Status

approved