cp's OEIS Frontend

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

%I A300295 #28 Nov 25 2024 05:01:03
%S A300295 1,15,105,63,99,429,195,85,969,133,483,1725,675,783,2697,1023,385,
%T A300295 3885,481,1599,5289,1935,2115,6909,2499,901,8745,1045,3363,10797,3843,
%U A300295 4095,13065,4623,1633,15549,1825,5775,18249,6399,6723,21165,7395,2581,24297,2821,8835,27645,9603,9999,31209,10815,3745
%N A300295 Denominator of (1/3)*n*(n + 2)/((1 + 2*n)*(3 + 2*n)).
%C A300295 The corresponding numerators are given in A144454(n+1).
%C A300295 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.
%C A300295 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.
%D A300295 L. B. W. Jolley, Summation of Series, Dover Publications, 2nd rev. ed., 1961, pp. 38, 40, 41.
%H A300295 Harvey P. Dale, <a href="/A300295/b300295.txt">Table of n, a(n) for n = 0..1000</a>
%H A300295 <a href="/index/Rec#order_27">Index entries for linear recurrences with constant coefficients</a>, 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).
%F A300295 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.
%F A300295 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))
%F A300295   = ((-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))).
%e A300295 The series begins: 1/(1*3*5) + 1/(3*5*7) + 1/(5*7*9) + ...
%e A300295 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,...
%t A300295 Table[(n(n+2))/(3(1+2n)(3+2n)),{n,0,60}]//Denominator (* _Harvey P. Dale_, Jun 19 2021 *)
%o A300295 (PARI) a(n) = denominator((1/3)*n*(n + 2)/((1 + 2*n)*(3 + 2*n))); \\ _Michel Marcus_, Mar 15 2018
%Y A300295 Cf. A005408, A144454(n+1) (numerators).
%K A300295 nonn,frac,easy
%O A300295 0,2
%A A300295 _Wolfdieter Lang_, Mar 15 2018