A230339 Numerator of Sum_{k=1..n} 1/(k(k+1)(k+2)(k+3)) = Sum_{k=1..n} 1/Pochhammer(k,4).

0, 1, 1, 19, 17, 55, 83, 119, 82, 73, 95, 121, 227, 559, 679, 815, 484, 1139, 443, 171, 295, 2023, 2299, 2599, 1462, 3275, 3653, 451, 749, 551, 5455, 5983, 3272, 7139, 7769, 8435, 1523, 3293, 3553, 11479, 6170, 13243, 14189, 15179, 8107, 5765

Offset: 0

Author

Jean-François Alcover, Oct 16 2013

Keywords

• nonn
• frac
• easy

Examples

			1/(1*2*3*4) + 1/(2*3*4*5) + 1/(3*4*5*6) = 19/360, so a(3) = 19.
The rationals r(n) = a(n)/A230340(n) begin: 0, 1/24, 1/20, 19/360, 17/315, 55/1008, 83/1512, 119/2160, 82/1485, 73/1320, 95/1716, 121/2184, 227/4095, 559/10080, 679/12240, 815/14688, ... - _Wolfdieter Lang_, Mar 08 2018
		

References

• L. B. W. Jolley, Summation of Series, Second revised ed., Dover, 1961, p.38, (202) and (201).

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Eric Weisstein's MathWorld, Pochhammer Symbol

Crossrefs

Cf. A001563, A052762, A094258, A125650, A230328, A230340 (denominators).

Programs

Mathematica

a[n_] := Numerator[1/18 - 1/(3*(n+1)*(n+2)*(n+3))]; Table[a[n], {n, 0, 100}]

PARI

a(n) = numerator(1/18 - 1/(3*(n+1)*(n+2)*(n+3))) \\ Colin Barker, Jul 30 2019

Formula

Numerator(1/18 - 1/(3*(n+1)*(n+2)*(n+3))) (from the generic formula Sum_{k=1..n} 1/Pochhammer(k, m) = 1/((m-1)*(m-1)!) - 1/((m-1)*Pochhammer(n+1, m-1)) with m = 4).

G.f. for the rationals r(n) = (1/18)*n*(11+n^2+6*n)/((1+n)*(n+2)*(n+3)) = a(n)/A230340(n): (1/18)*(1 - hypergeometric([1, 3], [4], -x/(1-x)))/(1-x) = (6*x - 15*x^2 + 11*x^3 + 6*(1 - 3*x + 3*x^2 - x^3)*log(1-x))/(36*x^3*(1-x)). - Wolfdieter Lang, Mar 08 2018

a(n) = numerator(1/18 - 1/(3*(n+1)*(n+2)*(n+3))). - Colin Barker, Jul 30 2019