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a(n) = floor((4th elementary symmetric function of S(n))/(3rd elementary symmetric function of S(n))), where S(n) = {first n+3 positive integers congruent to 1 mod 3}.
1

%I #25 Sep 08 2022 08:44:48

%S 0,2,4,7,11,16,22,28,35,43,51,60,70,81,93,105,118,132,146,161,177,194,

%T 212,230,249,269,289,310,332,355,379,403,428,454,480,507,535,564,594,

%U 624,655,687,719,752,786,821,857,893,930,968,1006,1045,1085,1126,1168,1210,1253,1297,1341,1386,1432

%N a(n) = floor((4th elementary symmetric function of S(n))/(3rd elementary symmetric function of S(n))), where S(n) = {first n+3 positive integers congruent to 1 mod 3}.

%H Robert Israel, <a href="/A024224/b024224.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (3,-4,4,-4,4,-4,4,-3,1).

%F G.f.: x^2*(2-2*x+3*x^2-2*x^3+3*x^4-2*x^5+2*x^6-x^7) / ((1-x)^3*(1+x^2)*(1+x^4)). - _Colin Barker_, Dec 10 2015

%F From _Robert Israel_, Dec 10 2015: (Start)

%F a(n) = floor(A024214(n+1)/A024213(n+1)).

%F a(n) = floor((3 n^2 + 5 n - 6)/8).

%F a(8*k+j) = 24*k^2 + (5 + 6*j) k + b(j), where b(j) = -1,0,2,4,7,11,16,22 for j = 0..7. (End)

%p seq(floor((3*n^2 + 5*n - 6)/8), n=1..100); # _Robert Israel_, Dec 10 2015

%t S[n_] := 3 Range[0, n + 2] + 1; Table[Floor[SymmetricPolynomial[4, S@ n]/SymmetricPolynomial[3, S@ n]], {n, 61}] (* _Michael De Vlieger_, Dec 10 2015 *)

%o (PARI) concat(0, Vec(x^2*(2-2*x+3*x^2-2*x^3+3*x^4-2*x^5+2*x^6-x^7)/((1-x)^3*(1+x^2)*(1+x^4)) + O(x^100))) \\ _Colin Barker_, Dec 10 2015

%o (PARI) a(n) = (3*n^2 + 5*n - 6)\8; \\ _Altug Alkan_, Dec 10 2015

%o (Magma) [(3*n^2+5*n-6) div 8: n in [1..70]]; // _Vincenzo Librandi_, Dec 11 2015

%Y Cf. A042413, A042414.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_

%E More terms from _Michael De Vlieger_, Dec 10 2015