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%I #11 Apr 05 2018 13:38:06
%S 0,1,11,21,17,25,69,91,29,9,175,209,123,143,329,375,53,119,531,589,
%T 325,357,781,851,231,125,1079,1161,623,667,1425,1519,101,429,1819,
%U 1925,1017,1073,2261,2379,625,164,2751,2881,1507
%N Numerators of n*(5 + 3*n)/(8*(1 + 3*n)*(4 + 3*n)), n >= 0.
%C The denominators are given in A300297.
%C The rational r(n) = a(n)/A300297(n) = (1/8)*n*(5 + 3*n)/(A(n)*A(n+1)) = 1/24 -(1/6)/(A(n)*A(n+1)), with the arithmetic progression A(k) = 1 + 3*k = A016777(k) is the value of the sum Sum_{k=0..n-1} 1/(A(k)*A(k+1)*A(k+2), for n >= 1, and r(0) = 0. See the Jolley reference, pp. 40-41, (210), and the general remark (201) on p. 38.
%D L. B. W. Jolley, Summation of Series, Dover Publications, 2nd rev. ed., 1961, pp. 38, 40, 41.
%F a(n) = numerator(r(n)), with r(n) = n*(5 + 3*n)/(8*(1 + 3*n)*(4 + 3*n)).
%F a(n) = n*(5 + 3*n)/32 if n == 0 or 9 (mod 32), a(n) = n*(5 + 3*n)/16 if n == 16 or 25 (mod 32), a(n) = n*(5 + 3*n)/8 if n == 1 or 8 or 17 or 24 (mod 32), and for other n one has a(n) = n*(5 + 3*n)/4 if n == 0 or 1 (mod 4) and a(n) = n*(5 + 3*n)/2 if n == 2 or 3 (mod 4).
%F G.f.: G(x) = (1/24)*(1 - hypergeometric([1, 2], [7/3], -x/(1-x)))/(1-x).
%e The series begins: 1/(1*4*7) + 1/(4*7*10) + 1/(7*10*13) + ...
%e The partial sums are r(n) = a(n)/A300297(n), n >= 1, and with r(0) = 0 they begin with 0/1, 1/28, 11/280, 21/520, 17/416, 25/608, 69/1672, 91/2200, 29/700, 9/217, 175/4216, 209/5032, 123/2960, 143/3440, 329/7912, 375/9016, ...
%o (PARI) a(n) = numerator(n*(5 + 3*n)/(8*(1 + 3*n)*(4 + 3*n))); \\ _Altug Alkan_, Mar 18 2018
%Y Cf. A016777, A300297.
%K nonn,frac,easy
%O 0,3
%A _Wolfdieter Lang_, Mar 16 2018
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