A210453 Decimal expansion of Sum_{n>=1} 1/(n*binomial(3*n,n)).

3, 7, 1, 2, 1, 6, 9, 7, 5, 2, 6, 0, 2, 4, 7, 0, 3, 4, 4, 7, 4, 7, 7, 1, 6, 6, 6, 0, 7, 5, 3, 5, 8, 8, 0, 5, 5, 8, 7, 6, 2, 9, 4, 6, 9, 0, 5, 1, 9, 7, 2, 2, 2, 1, 3, 6, 4, 7, 7, 8, 9, 3, 9, 5, 7, 3, 4, 0, 0, 0, 8, 3, 5, 3, 5, 5, 9, 8, 4, 9, 6, 9, 1, 3, 1, 4, 3, 2, 7, 5, 4, 1, 7, 7, 6, 5, 0, 5, 0, 9, 9, 2, 3, 2, 3, 9, 6, 1, 7, 5, 6, 9, 0, 7, 7, 3, 5, 3, 5, 2, 7, 3, 1, 6, 8, 6

Offset: 0

Author

R. J. Mathar, Jan 21 2013

Keywords

• cons
• nonn

Examples

			0.37121697526024703447477166607535880558762946905197...
		

References

• George Boros and Victor H. Moll, Irresistible integrals, Cambridge University Press, 2006, p. 60.

Links

• Necdet Batir, On the series Sum_{k=1..oo} binomial(3k,k)^{-1} k^{-n} x^k, Proc. Indian Acad. Sci. (Math. Sci.), Vol. 115, No. 4 (2005), pp. 371-381; arXiv preprint, arXiv:math/0512310 [math.AC], 2005.
• Courtney Moen, Infinite series with binomial coefficients, Math. Mag. 64 (1) (1991) 53-55.

Crossrefs

Cf. A005809, A073010, A075778, A210462, A210463.

Programs

Maple

A075778neg := proc()
        1/3-root[3](25/2-3*sqrt(69)/2)/3 -root[3](25/2+3*sqrt(69)/2)/3;
end proc:
A210462 := proc()
        local a075778 ;
        a075778 := A075778neg() ;
        (1+1/a075778/(a075778-1))/2 ;
end proc:
A210463 := proc()
        local a075778,a210462 ;
        a075778 := A075778neg() ;
        a210462 := A210462() ;
        -1/a075778-a210462^2 ;
        sqrt(%) ;
end proc:
A210453 := proc()
        local v,x;
        v := 0.0 ;
        for x in [ A075778neg(), A210462()+I*A210463(), A210462()-I*A210463() ] do
                v := v+ x*log(1-1/x)/(3*x-2) ;
        end do:
        evalf(v) ;
end proc:
A210453() ;

Mathematica

RealDigits[ HypergeometricPFQ[{1, 1, 3/2}, {4/3, 5/3}, 4/27]/3, 10, 105] // First (* Jean-François Alcover, Feb 11 2013 *)

Formula

Equals Sum_{n>=1} 1/(n*A005809(n)).

Equals Integral_{x=0..1} x^2/(1-x^2+x^3) dx.

Equals Sum_(R) R*log(1-1/R)/(3*R-2) where R is summed over the set of the three constants -A075778, A210462-i*A210463 and A210462-i*A210463, i=sqrt(-1), that is, over the set of the three roots of x^3-x^2+1.

Equals (1/sqrt(23)) * (arctan(sqrt(3)/(2*phi-1)) * 18*phi/(phi^2-phi+1) - log((phi^3+1)/(phi+1)^3) * (3*sqrt(3)*phi*(1-phi))/(phi^3+1)), where phi = ((25+3*sqrt(69))/2)^(1/3) (Batir, 2005, p. 378, eq. (3.2)). - Amiram Eldar, Dec 07 2024