A024183 Second elementary symmetric function of 3,4,...,n+3.

12, 47, 119, 245, 445, 742, 1162, 1734, 2490, 3465, 4697, 6227, 8099, 10360, 13060, 16252, 19992, 24339, 29355, 35105, 41657, 49082, 57454, 66850, 77350, 89037, 101997, 116319, 132095, 149420, 168392, 189112, 211684, 236215, 262815, 291597, 322677, 356174

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• easy

Links

• Ivan Neretin, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Programs

GAP

List([1..40],n->n*(n+1)*(3*n^2+35*n+106)/24); # Muniru A Asiru, May 19 2018

Magma

[n*(n+1)*(3*n^2+35*n+106)/24: n in [1..40]]; // Vincenzo Librandi, May 03 2018

Maple

seq(n*(n+1)*(3*n^2+35*n+106)/24,n=1..40); # Muniru A Asiru, May 19 2018

Mathematica

f[k_] := k + 2; t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[2, t[n]]
Table[a[n], {n, 2, 30}]  (* A024183 *)
(* Clark Kimberling, Dec 31 2011 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {12, 47, 119, 245, 445}, 40] (* Vincenzo Librandi, May 03 2018 *)

PARI

Vec(-x*(4*x^2-13*x+12)/(x-1)^5 + O(x^100)) \\ Colin Barker, Aug 15 2014

Formula

a(n) = n*(n+1)*(3*n^2 + 35*n + 106)/24.

If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k) * Stirling1(n-k,i) * Product_{j=0..k-1} (-a-j), then a(n-2) = f(n,n-2,3), for n >= 3. - Milan Janjic, Dec 20 2008

From Colin Barker, Aug 15 2014: (Start)

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).

G.f.: -x*(4*x^2-13*x+12)/(x-1)^5. (End)

E.g.f.: exp(x)*x*(6 + x)*(48 + 38*x + 3*x^2)/24. - Elmo R. Oliveira, Aug 15 2025