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.

A181355 a(3*n+1) = 4^(2^n), a(3*n+2) = 3^(2^n), a(3*n+3) = 4^(2^n) - 3^(2^n).

Original entry on oeis.org

4, 3, 1, 16, 9, 7, 256, 81, 175, 65536, 6561, 58975, 4294967296, 43046721, 4251920575, 18446744073709551616, 1853020188851841, 18444891053520699775, 340282366920938463463374607431768211456, 3433683820292512484657849089281
Offset: 1

Views

Author

Jamel Ghanouchi, Jan 27 2011

Keywords

Comments

Previous name was: Consider pairs of fractions (x,y) starting (4,3) and updated via z:=1/(1/x+1/y), x->x-z, y->y-z. The sequence shows the triples (numerator(x), numerator(y), numerator(x)-numerator(y)) after each update.

Examples

			(x=4,y=3) is shown as the first triple (4,3,1) in the sequence. This generates z=12/7 which generates the new pair (x,y) = (16/7,9/7) shown as (16,9,7). - _R. J. Mathar_, Feb 09 2011
		

Programs

  • Maple
    x := 4 ; y := 3 ;
    for loo from 1 to 7 do printf("%d, %d, %d, ", numer(x), numer(y), numer(x)-numer(y)) ; z := 1/(1/x+1/y) ; x := x-z ; y := y-z ; end do: # R. J. Mathar, Feb 09 2011

Formula

a(3*n+1) = 4^(2^n), a(3*n+2) = 3^(2^n), a(3*n+3) = 4^(2^n) - 3^(2^n). - Philippe Deléham , Oct 29 2013

Extensions

Corrected by Philippe Deléham, Oct 29 2013
New name using Philippe Deléham's formula, Joerg Arndt, Nov 14 2014