A024219 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}.

0, 3, 7, 12, 19, 28, 38, 49, 62, 77, 93, 110, 129, 150, 172, 195, 220, 247, 275, 304, 335, 368, 402, 437, 474, 513, 553, 594, 637, 682, 728, 775, 824, 875, 927, 980, 1035, 1092, 1150, 1209, 1270, 1333, 1397, 1462, 1529, 1598, 1668, 1739, 1812, 1887, 1963, 2040

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• easy

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1000
• Wikipedia, Elementary symmetric polynomial
• Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Crossrefs

Cf. A000326, A024212,

Programs

Mathematica

LinearRecurrence[{3,-4,4,-3,1},{0,3,7,12,19},60] (* Harvey P. Dale, May 20 2019 *)

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

PARI

a(n) = floor(n*(9*n^2+9*n-2)/(4*(3*n+2))); \\ Andrew Howroyd, Aug 12 2018

Formula

From R. J. Mathar, Oct 08 2011: (Start)

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

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

From Andrew Howroyd, Aug 12 2018: (Start)

The above conjectures are true.

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

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

(End)

Extensions

More terms from Joshua Zucker, May 20 2006