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.

User: Maurice Mischler

Maurice Mischler's wiki page.

Maurice Mischler has authored 2 sequences.

A086208 Prime numbers p with the property that the sequence p-1*2, p-2*3, p-3*4, p-4*5, ..., consists only of primes and a final 1.

Original entry on oeis.org

3, 7, 13, 43, 73
Offset: 0

Author

Maurice Mischler (maurice.mischler(AT)ima.unil.ch), Aug 27 2003

Keywords

Comments

Related to the fact that the list 5, 8, 13, 29, 53, 173, 293 gives the only positive discriminants of class number 1 of the form j^2+1.
If we allow the first number that we subtract to be greater than 2, the next sequence after 73, 71, 67, 61,... is 2003, 1831, 1657, 1481, 1303, 1123, 941, 757, 571, 383, 193, 1: start with 2003 and subtract 172,174,176, etc., which is an amusing property of 2003.

Examples

			For example 73 71 67 61 53 43 31 17 1. If the GRH is true this is the last example.
		

Extensions

Definition corrected by Harvey P. Dale, Jul 25 2015

A086695 a(n) = 100^n - 10^n - 1.

Original entry on oeis.org

89, 9899, 998999, 99989999, 9999899999, 999998999999, 99999989999999, 9999999899999999, 999999998999999999, 99999999989999999999, 9999999999899999999999, 999999999998999999999999, 99999999999989999999999999, 9999999999999899999999999999
Offset: 1

Author

Maurice Mischler (maurice.mischler(AT)ima.unil.ch), Sep 12 2003

Keywords

Comments

Digits of the inverses of these numbers give the Fibonacci numbers. More precisely, the digits of 1/(10^(2*n)-10^n-1) give the Fibonacci numbers up to 10^n.
More generally, if x_1, x_2, x_n=x_(n-1)-x_(n-2) is any Lucas sequence, then the digits of the numbers (x_1*10^n-(x_1-x_2))/(10^(2*n)-10^n-1) give the x_n up to 10^n.
1/a(n) = Sum_{i>=1} A000045(i-1)/10^(n*i) (see Long paper). - Michel Marcus, May 01 2013

Crossrefs

Cf. A000045.

Programs

  • Mathematica
    Table[100^n-10^n-1,{n,20}] (* or *) LinearRecurrence[{111,-1110,1000},{89,9899,998999},20] (* Harvey P. Dale, Nov 16 2023 *)
  • PARI
    a(n)=100^n-10^n-1 \\ Charles R Greathouse IV, May 01 2013

Formula

a(n) = 10^(2*n) - 10^n - 1.

Extensions

Offset corrected by Jon E. Schoenfield, Jun 17 2018