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.

A357502 a(n) = ((1 + sqrt(n))^n - (1 - sqrt(n))^n)/(2*sqrt(n)).

Original entry on oeis.org

1, 2, 6, 20, 80, 342, 1624, 8136, 43776, 246410, 1463264, 9033180, 58200064, 387905182, 2679200640, 19068105488, 139929124864, 1054773505170, 8167509816832, 64795371984420, 526534098026496, 4374163243287398, 37135913476691968, 321727849480560600
Offset: 1

Views

Author

Alexander R. Povolotsky, Oct 01 2022

Keywords

Crossrefs

Main diagonal of A099173.

Programs

  • Maple
    b:= proc(n, k) option remember;
     `if`(n<2, n, 2*b(n-1, k)+(k-1)*b(n-2, k))
    end:
a:= n-> b(n$2):
seq(a(n), n=1..26);  # Alois P. Heinz, Oct 01 2022
  • Mathematica
    Simplify[Table[((1 + Sqrt[n])^n - (1 - Sqrt[n])^n)/(2*Sqrt[n]),{n,26}]] (* Stefano Spezia, Oct 03 2022 *)
  • PARI
    a(n) = polcoeff(lift( Mod('x,'x^2-2*'x+1-n)^n ),1); \\ Kevin Ryde, Oct 01 2022
  • Python
    from sympy import simplify, sqrt
def A357502(n): return simplify(((1+sqrt(n))**n-(1-sqrt(n))**n)/sqrt(n))>>1 # Chai Wah Wu, Oct 14 2022

Formula

From Alois P. Heinz, Oct 01 2022: (Start)
a(n) = [x^n] x/(1-2*x-(n-1)*x^2).
a(n) = Sum_{j=0..floor(n/2)} n^j * binomial(n,2*j+1).
a(n) = A099173(n,n). (End)

A357580 a(n) = ((1 + sqrt(n))^n - (1 - sqrt(n))^n)/(2*n*sqrt(n)).

Original entry on oeis.org

1, 1, 2, 5, 16, 57, 232, 1017, 4864, 24641, 133024, 752765, 4476928, 27707513, 178613376, 1191756593, 8231124992, 58598528065, 429868937728, 3239768599221, 25073052286976, 198825601967609, 1614604933769216, 13405327061690025, 113725655719346176
Offset: 1

Views

Author

Alexander R. Povolotsky, Oct 04 2022

Keywords

Links

Crossrefs

Cf. A099173, A357502.

Programs

  • Maple
    b:= proc(n, k) option remember;
     `if`(n<2, n, 2*b(n-1, k)+(k-1)*b(n-2, k))
    end:
a:= n-> b(n$2)/n:
seq(a(n), n=1..25);  # Alois P. Heinz, Oct 04 2022
  • Mathematica
    Expand[Table[((1 + Sqrt[n])^n - (1 - Sqrt[n])^n)/(2*n*Sqrt[n]), {n, 1, 27}]]
  • Python
    from sympy import simplify, sqrt
def A357580(n): return simplify(((1+sqrt(n))**n-(1-sqrt(n))**n)/(n*sqrt(n)))>>1 # Chai Wah Wu, Oct 14 2022

Formula

a(n) = A357502(n)/n.
From Alois P. Heinz, Oct 04 2022: (Start)
a(n) = [x^n] x/(n*(1-2*x-(n-1)*x^2)).
a(n) = Sum_{j=0..floor(n/2)} n^(j-1) * binomial(n,2*j+1).
a(n) = A099173(n,n)/n. (End)
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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