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.

A342205 a(n) = T(n,n+1) where T(n,x) is a Chebyshev polynomial of the first kind.

Original entry on oeis.org

1, 2, 17, 244, 4801, 120126, 3650401, 130576328, 5374978561, 250283080090, 13007560326001, 746411226303612, 46873096812360001, 3197490648645613334, 235451028081583642049, 18614381236112230383376, 1572584048032918633353217
Offset: 0

Views

Author

Seiichi Manyama, Mar 05 2021

Keywords

Crossrefs

Programs

  • Mathematica
    Table[ChebyshevT[n, n + 1], {n, 0, 16}] (* Amiram Eldar, Mar 05 2021 *)
  • PARI
    a(n) = polchebyshev(n, 1, n+1);
    
  • PARI
    a(n) = round(cos(n*acos(n+1)));
    
  • PARI
    a(n) = if(n==0, 1, n*sum(k=0, n, (2*n)^k*binomial(n+k, 2*k)/(n+k)));

Formula

a(n) = cos(n*arccos(n+1)).
a(n) = n * Sum_{k = 0..n} (2*n)^k * binomial(n+k,2*k)/(n+k) for n > 0.
From Peter Bala, Mar 11 2024: (Start)
a(2*n+1) == 1 (mod (2*n + 1)^3); a(2*n) == 1 (mod (n + 1)*(2*n)^3).
a(n) = hypergeom([n, -n], [1/2], -n/2). (End)
a(n) ~ exp(1) * 2^(n-1) * n^n. - Vaclav Kotesovec, Mar 12 2024

A343261 a(n) = 2 * T(n,(n+2)/2) where T(n,x) is a Chebyshev polynomial of the first kind.

Original entry on oeis.org

2, 3, 14, 110, 1154, 15127, 238142, 4379769, 92198402, 2186871698, 57721023502, 1678243366813, 53301709843202, 1836220544383695, 68200709735854334, 2716906424134261502, 115561578124838522882, 5227260815326346060059, 250566480717349417632398
Offset: 0

Views

Author

Seiichi Manyama, Apr 09 2021

Keywords

Crossrefs

Main diagonal of A299741.

Programs

  • Mathematica
    Table[2*ChebyshevT[n, (n+2)/2], {n, 0, 18}] (* Amiram Eldar, Apr 09 2021 *)
  • PARI
    a(n) = 2*polchebyshev(n, 1, (n+2)/2);
    
  • PARI
    a(n) = round(2*cos(n*acos((n+2)/2)));
    
  • PARI
    a(n) = if(n==0, 2, 2*n*sum(k=0, n, n^k*binomial(n+k, 2*k)/(n+k)));

Formula

a(n) = 2 * cos(n*arccos((n+2)/2)).
a(n) = 2 * n * Sum_{k=0..n} n^k * binomial(n+k,2*k)/(n+k) for n > 0.
a(n) ~ exp(2) * n^n. - Vaclav Kotesovec, Apr 09 2021
Showing 1-2 of 2 results.