cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A024196 a(n) = 2nd elementary symmetric function of the first n+1 odd positive integers.

Original entry on oeis.org

3, 23, 86, 230, 505, 973, 1708, 2796, 4335, 6435, 9218, 12818, 17381, 23065, 30040, 38488, 48603, 60591, 74670, 91070, 110033, 131813, 156676, 184900, 216775, 252603, 292698, 337386, 387005, 441905, 502448, 569008, 641971, 721735, 808710, 903318
Offset: 1

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Examples

			a(8) = 8*80+7*79+6*78+5*77+4*76+3*75+2*74+1*73 = 2796. - _Bruno Berselli_, Mar 13 2012
		

Crossrefs

From Johannes W. Meijer, Jun 08 2009: (Start)
Equals third right hand column of A028338 triangle.
Equals third left hand column of A109692 triangle.
Equals third right hand column of A161198 triangle divided by 2^m.
(End)
Cf. A016061.

Programs

  • GAP
    List([1..36], n -> n*(n+1)*(3*n^2+5*n+1)/6); # Muniru A Asiru, Feb 13 2018
  • Maple
    seq(n*(n+1)*(3*n^2+5*n+1)/6,n=1..25); # Muniru A Asiru, Feb 13 2018
  • Mathematica
    f[k_] := 2 k - 1; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[2, t[n]]
    Table[a[n], {n, 2, 50}]  (* A024196 *)
    (* Clark Kimberling, Dec 31 2011 *)
    Table[(n(n+1)(3n^2+5n+1))/6,{n,50}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{3,23,86,230,505},50] (* Harvey P. Dale, Jul 08 2019 *)

Formula

a(n) = n*(n+1)*(3*n^2+5*n+1)/6.
From Bruno Berselli, Mar 13 2012: (Start)
G.f.: x*(3 + 8*x + x^2)/(1 - x)^5.
a(n) = Sum_{i=1..n} (n+1-i)*((n+1)^2-i).
a(n) = n*A016061(n) - Sum_{i=0..n-1} A016061(i). (End)
a(n) - a(n-1) = A099721(n). Partial sums of A099721.- Philippe Deléham, May 07 2012
a(n) = Sum_{i=1..n} ((2*i-1)*Sum_{j=i..n} (2*j+1)) = 1*(3+5+...2*n+1) + 3*(5+7+...+2*n+1) + ... + (2*n-1)*(2*n+1). - J. M. Bergot, Apr 21 2017
a(n) = A028338(n+1, n-1), n >= 1, (third diagonal). See the crossref. below. Wolfdieter Lang, Jul 21 2017
a(n) = (A000583(n+1) - A000447(n+1))/2. - J. M. Bergot, Feb 13 2018