A024219 - OEIS

%I #29 May 20 2019 10:26:13

%S 0,3,7,12,19,28,38,49,62,77,93,110,129,150,172,195,220,247,275,304,

%T 335,368,402,437,474,513,553,594,637,682,728,775,824,875,927,980,1035,

%U 1092,1150,1209,1270,1333,1397,1462,1529,1598,1668,1739,1812,1887,1963,2040

%N a(n) = floor( (2nd elementary symmetric function of S(n))/(first elementary symmetric function of S(n)) ), where S(n) = {first n+1 positive integers congruent to 1 mod 3}.

%H Andrew Howroyd, <a href="/A024219/b024219.txt">Table of n, a(n) for n = 1..1000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial">Elementary symmetric polynomial</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,-4,4,-3,1).

%F From _R. J. Mathar_, Oct 08 2011: (Start)

%F Conjecture: a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5);

%F g.f.: x^2*(-3+2*x-3*x^2+x^3) / ( (x^2+1)*(x-1)^3 ). (End)

%F From _Andrew Howroyd_, Aug 12 2018: (Start)

%F The above conjectures are true.

%F a(n) = floor(A024212(n) / A000326(n+1)).

%F a(n) = floor(n*(9*n^2 + 9*n - 2)/(4*(3*n + 2))).

%F (End)

%t LinearRecurrence[{3,-4,4,-3,1},{0,3,7,12,19},60] (* _Harvey P. Dale_, May 20 2019 *)

%o (PARI) a(n)=floor(sum(j=0, n, sum(k=j+1, n, (3*j+1)*(3*k+1)))/sum(i=0, n, (3*i+1))) \\ _Andrew Howroyd_, Aug 12 2018

%o (PARI) a(n) = floor(n*(9*n^2+9*n-2)/(4*(3*n+2))); \\ _Andrew Howroyd_, Aug 12 2018

%Y Cf. A000326, A024212,

%K nonn,easy

%O 1,2

%A _Clark Kimberling_

%E More terms from _Joshua Zucker_, May 20 2006