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.

Showing 1-2 of 2 results.

A203246 Second elementary symmetric function of the first n terms of (1,1,2,2,3,3,4,4,...).

Original entry on oeis.org

1, 5, 13, 31, 58, 106, 170, 270, 395, 575, 791, 1085, 1428, 1876, 2388, 3036, 3765, 4665, 5665, 6875, 8206, 9790, 11518, 13546, 15743, 18291, 21035, 24185, 27560, 31400, 35496, 40120, 45033, 50541, 56373, 62871, 69730, 77330, 85330, 94150, 103411, 113575
Offset: 2

Views

Author

Clark Kimberling, Dec 31 2011

Keywords

Comments

Second subdiagonal of A246117. - Peter Bala, Aug 15 2014

Crossrefs

Cf. A203298, A203299, A246117, A212523 (odd bisection), A103220 (even bisection).

Programs

  • Mathematica
    f[k_] := Floor[(k + 1)/2]; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[2, t[n]]
    Table[a[n], {n, 2, 50}]  (* A203246 *)

Formula

Conjectural o.g.f.: x^2*(1 + 3*x + x^2 + x^3)/((1 - x^2)^3*(1 - x)^2). - Peter Bala, Aug 15 2014
Conjectural closed form: 64*a(n) = 2*n^2 -16*n/3 -3 +16*n^3/3 +2*n^4 +(-1)^n *(3-2*n^2). - R. J. Mathar, Oct 01 2016
Both conjectures are true. See link. - Sela Fried, Dec 22 2024

A203298 Second elementary symmetric function of the first n terms of (1,2,2,3,3,4,4,5,5...).

Original entry on oeis.org

2, 8, 23, 47, 91, 151, 246, 366, 540, 750, 1037, 1373, 1813, 2317, 2956, 3676, 4566, 5556, 6755, 8075, 9647, 11363, 13378, 15562, 18096, 20826, 23961, 27321, 31145, 35225, 39832, 44728, 50218, 56032, 62511, 69351, 76931, 84911, 93710, 102950
Offset: 2

Views

Author

Clark Kimberling, Dec 31 2011

Keywords

Crossrefs

Programs

  • Mathematica
    f[k_] := Floor[(k + 2)/2]; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[2, t[n]]
    Table[a[n], {n, 2, 50}]  (* A203298 *)

Formula

Empirical g.f.: x^2*(2 + 4*x + 3*x^2 - 3*x^3 - x^4 + x^5) / ((1 - x)^5*(1 + x)^3). - Colin Barker, Aug 15 2014
Conjectures from Colin Barker, Jan 04 2018: (Start)
a(n) = (6*n^4 + 40*n^3 + 48*n^2 - 112*n) / 192 for n even.
a(n) = (6*n^4 + 40*n^3 + 36*n^2 - 136*n + 54) / 192 for n odd.
a(n) = 2*a(n-1) + 2*a(n-2) - 6*a(n-3) + 6*a(n-5) - 2*a(n-6) - 2*a(n-7) + a(n-8) for n>9.
(End)
Showing 1-2 of 2 results.