A046825 Numerator of Sum_{k=0..n} 1/binomial(n,k).

1, 2, 5, 8, 8, 13, 151, 256, 83, 146, 1433, 2588, 15341, 28211, 52235, 19456, 19345, 36362, 651745, 6168632, 1463914, 2786599, 122289917, 233836352, 140001721, 268709146, 774885169, 1491969394, 41711914513, 80530073893

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• frac
• nice

Comments

The term a(12)=15341 is divisible by 23^2. Is there another term a(n) divisible by the square of a prime p larger than n+1? - M. F. Hasler, Jul 17 2012

Examples

			1, 2, 5/2, 8/3, 8/3, 13/5, 151/60, 256/105, 83/35, 146/63, 1433/630, 2588/1155, 15341/6930, 28211/12870, 52235/24024, 19456/9009, 19345/9009, ... = A046825/A046826
		

References

• L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 294, Problem 7.15.
• R. L. Graham, D. E. Knuth, O. Patashnik; Concrete Mathematics, Addison-Wesley, Reading (1994) 2nd Ed. Exercise 5.100.
• G. Letac, Problèmes de probabilités, Presses Universitaires de France (1970), p. 14.
• F. Nedemeyer and Y. Smorodinsky, Resistances in the multidimensional cube, Quantum 7:1 (1996) 12-15 and 63.

Links

• T. D. Noe, Table of n, a(n) for n = 0..200
• T. Mansour, Gamma function, beta function and combinatorial identities.
• T. Sillke, More information
• D. Singmaster, Problem 79-16, Resistances in an n-Dimensional Cube, SIAM Review, 22 (1980) 504.
• B. Sury, Sum of the reciprocals of the binomial coefficients, Europ. J. Combinatorics, 14 (1993), 351-353.

Crossrefs

Cf. A003149, A046826, A048211, A051389, A100512, A100513.

Programs

Magma

[Numerator((&+[1/Binomial(n,j): j in [0..n]])): n in [0..40]]; // G. C. Greubel, May 24 2021

Mathematica

Numerator/@Table[Sum[1/Binomial[n,k],{k,0,n}],{n,0,40}]  (* Harvey P. Dale, Apr 21 2011 *)

PARI

P=1;vector(30,n,numerator(P)+0*P=P/2/n*(n+1)+1) \\ M. F. Hasler, Jul 17 2012

PARI

A046825(n)=numerator(sum(k=0,n,1/binomial(n,k))) \\ M. F. Hasler, Jul 19 2012

Sage

[numerator(sum(1/binomial(n,j) for j in (0..n))) for n in (0..40)] # G. C. Greubel, May 24 2021

Formula

Let P(n) = (1/n) * Sum_{k=0..n-1} 1/binomial(n-1, k) = A046878(n)/A046879(n) = A046825(n-1)/(n*A046826(n-1)): { 0, 1, 1, 5/6, 2/3, 8/15, ...}. Then P(n) = 2^(-n) * Sum_{k=1..n} 2^k / k = 2^(-n+1) * Sum_{k odd} binomial(n, k)/k; P(0) = 0, P(n) = P(n-1)/2 + 1/n. - Torsten Sillke (Torsten.Sillke(AT)uni-bielefeld.de)

G.f. for P(n): (2*log(1-z))/(-2+z). - Wouter Meeussen

P(n) = 2^(-n) * Sum_{k=1..n} (binomial(n,k) + 1)/k.

a(n) = numerator( A003149(n)/n! ). - G. C. Greubel, May 24 2021

Extensions

References entries (Comtet, Graham et al., Letac, Nedemeyer) and Links entries (Singmaster, Sury) from Torsten.Sillke(AT)uni-bielefeld.de