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.

A242645 a(n) = concatenation of decimal expansions of powers of 11 (in decreasing order).

Original entry on oeis.org

1, 111, 121111, 1331121111, 146411331121111, 161051146411331121111, 1771561161051146411331121111, 194871711771561161051146411331121111, 214358881194871711771561161051146411331121111, 2357947691214358881194871711771561161051146411331121111
Offset: 0

Views

Author

N. J. A. Sloane, May 23 2014

Keywords

Comments

A068704(11) (presently unknown) is either 0, if every term here is composite, or the index of the first prime in the sequence.

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember;
          `if`(n=0, 1, parse(cat(11^n, a(n-1))))
        end:
    seq(a(n), n=0..10);  # Alois P. Heinz, May 24 2014

Formula

a(n) is the concatenation of 11^n, 11^(n-1), ..., 121, 11, 1.

A386552 Concatenate powers of 10.

Original entry on oeis.org

1, 110, 110100, 1101001000, 110100100010000, 110100100010000100000, 1101001000100001000001000000, 110100100010000100000100000010000000, 110100100010000100000100000010000000100000000, 1101001000100001000001000000100000001000000001000000000
Offset: 0

Views

Author

Jason Bard, Jul 25 2025

Keywords

Comments

Binary version of A045507. Base-2 representation of A164894.
Concatenate first A000217(n+1) terms of A010054.

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember;
          `if`(n<0, 0, parse(cat(a(n-1), 10^n)))
        end:
    seq(a(n), n=0..10);  # Alois P. Heinz, Jul 28 2025
  • Mathematica
    a[0] = 1; a[n_] := a[n - 1]*10^(n+1) + 10^n; Table[a[n], {n, 0, 9}]
  • Python
    def A386552(n): return 10**n*sum(10**(k*((n<<1)-k+1)>>1) for k in range(n+1)) # Chai Wah Wu, Aug 05 2025

Formula

a(n) = Sum_{k=1..n+1} 10^A133082(k,n+2).
a(n) = A101305(n) + 10^A000096(n).
For n >= 1, a(n) = 10^(n+1)*a(n-1)+10^n.
Number of digits in a(n) is A000217(n+1).
Showing 1-2 of 2 results.