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Numerator of n*(n+5)/((n+2)*(n+3)).
2

%I #36 Aug 11 2022 08:54:56

%S 0,1,7,4,6,25,11,14,52,21,25,88,34,39,133,50,56,187,69,76,250,91,99,

%T 322,116,125,403,144,154,493,175,186,592,209,221,700,246,259,817,286,

%U 300,943,329,344,1078,375,391,1222

%N Numerator of n*(n+5)/((n+2)*(n+3)).

%H Vincenzo Librandi, <a href="/A027625/b027625.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,3,0,0,-3,0,0,1).

%F G.f.: x*(1 + 7*x + 4*x^2 + 3*x^3 + 4*x^4 - x^5 - x^6 - 2*x^7)/(1 - x^3)^3.

%F a(n) = numerator of n*(n+5)/6. - _Altug Alkan_, Apr 18 2018

%F From _Peter Bala_, Aug 06 2022: (Start)

%F a(n) is quasi-polynomial in n:

%F a(3*n) = (1/2)*n*(3*n+5) = A115067(n+1).

%F a(3*n+1) = (1/2)*(n+2)*(3*n+1) = A095794(n+1).

%F a(3*n+2) = (1/2)*(3*n+2)*(3*n+7) = A179436(n). (End)

%F Sum_{n>=1} 1/a(n) = 4*Pi/(15*sqrt(3)) + 87/50. - _Amiram Eldar_, Aug 11 2022

%t CoefficientList[Series[x*(1+7*x+4*x^2+3*x^3+4*x^4-x^5-x^6-2*x^7)/(1-x^3)^3, {x, 0, 50}], x] (* _Vincenzo Librandi_, Mar 04 2014 *)

%t Numerator[25*Binomial[Range[0, 50]/5 +1, 2]/3] (* _G. C. Greubel_, Aug 05 2022 *)

%o (Magma) [Numerator(n*(n+5)/((n+2)*(n+3))): n in [0..50]]; // _Vincenzo Librandi_, Mar 04 2014

%o (PARI) a(n) = numerator(n*(n+5)/6); \\ _Altug Alkan_, Apr 18 2018

%o (SageMath) [numerator(n*(n+5)/6) for n in (0..50)] # _G. C. Greubel_, Aug 05 2022

%Y Cf. A027626 (denominator), A095794, A115067, A179436.

%K nonn,frac,easy

%O 0,3

%A _N. J. A. Sloane_