A300295 Denominator of (1/3)*n*(n + 2)/((1 + 2*n)*(3 + 2*n)).

1, 15, 105, 63, 99, 429, 195, 85, 969, 133, 483, 1725, 675, 783, 2697, 1023, 385, 3885, 481, 1599, 5289, 1935, 2115, 6909, 2499, 901, 8745, 1045, 3363, 10797, 3843, 4095, 13065, 4623, 1633, 15549, 1825, 5775, 18249, 6399, 6723, 21165, 7395, 2581, 24297, 2821, 8835, 27645, 9603, 9999, 31209, 10815, 3745

Offset: 0

Author

Wolfdieter Lang, Mar 15 2018

Keywords

• nonn
• frac
• easy

Comments

The corresponding numerators are given in A144454(n+1).

r(n) = A144454(n+1)/a(n) = Sum_{k=0..n-1} 1/(A(k)*A(k+1)*A(k+2)), with A(j) = 1 + 2*j = A005408(n) for n >= 1, and r(0) = 0. This can be written as r(n) = 1/12 - 1/(4*A(n)*A(n+1)) = (1/3)*n*(n + 2)/(A(n)*A(n+1)). See Jolley, p. 40/41, (209), and the general remark on p. 38, (201). The value of the infinite series is therefore 1/12.

For the proof that numerator(r(n)) = A144454(n+1) one checks the formula with (mod 9) and (mod 3) given there. E.g., if n = 1 + 9*k then r(n-1) = k*(2 + 9*k)/((1 + 6*k)*(1 + 18*k)) and numerator(r(n-1)) = k*(2 + 9*k) = ((n-1)^2 - )/9 as claimed, because this ratio for r(n-1) is in lowest terms.

Examples

			The series begins: 1/(1*3*5) + 1/(3*5*7) + 1/(5*7*9) + ...
The partial sums are r(n) = A144454(n+1)/a(n), n >= 1, and with r(0) = 0 they begin with 0/1, 1/15, 8/105, 5/63, 8/99, 35/429, 16/195, 7/85, 80/969, 11/133, 40/483, 143/1725, 56/675, 65/783, 224/2697, 85/1023, 32/385,...
		

References

• L. B. W. Jolley, Summation of Series, Dover Publications, 2nd rev. ed., 1961, pp. 38, 40, 41.

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,-3,0,0,0,0,0,0,0,0,1).

Crossrefs

Cf. A005408, A144454(n+1) (numerators).

Programs

Mathematica

Table[(n(n+2))/(3(1+2n)(3+2n)),{n,0,60}]//Denominator (* Harvey P. Dale, Jun 19 2021 *)

PARI

a(n) = denominator((1/3)*n*(n + 2)/((1 + 2*n)*(3 + 2*n))); \\ Michel Marcus, Mar 15 2018

Formula

a(n) = denominator(r(n)), with r(n) = (1/3)*n*(n + 2)/((1 + 2*n)*(3 + 2*n)), n >= 0. r(n-1) = (1/3)*(n^2 - 1)/((2*n)^2 -1), n >= 1.

G.f. for r(n) = A144454(n+1)/a(n): G(x) = (1/12)*(1 - hypergeometric([1, 2], [5/2], -x/(1-x)))/(1-x) = ((-3 + 5*x)*sqrt(x)/sqrt(1 - x) + 3*sqrt(1 - x)*(1 - x)*arcsinh(sqrt(x)/sqrt(1 - x)))/(24*x*(1 - x)*sqrt(x)/sqrt(1 - x))

= ((-3 + 5*x)*sqrt(x/(1-x)) + 3*(1 - x)*sqrt(1 - x)*log((1 + sqrt(x))/sqrt(1 - x)))/(24*x*(1 - x)*sqrt(x/(1 - x))).