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 1-4 of 4 results.

A051840 Integer part of the Verhulst sequence x(n)=x(n-1)+3*(1-x(n-1))*x(n-1), x(0)=.1.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0
Offset: 0

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Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 13 1999

Keywords

Comments

a(n) = floor(A220811(n)/A220812(n)). - Reinhard Zumkeller, Dec 22 2012

References

  • Peitgen, H.-O. and Richter, P. H., The Beauty of Fractals. Springer, 1986, p. 24.

Programs

  • Haskell
    a051840 n = a051840_list !! n
    a051840_list = map floor vs where
       vs = iterate (\x -> x * (4 - 3 * x)) 0.1
    -- Reinhard Zumkeller, Dec 22 2012
  • Mathematica
    NestList[ #*(4-3*#)&, 0.1, 103] // Floor (* Jean-François Alcover, Jan 16 2013 *)

Extensions

Data corrected for n > 37 by Reinhard Zumkeller, Dec 22 2012

A220811 Numerators of the Verhulst sequence x(n+1)=4*x(n)-3*x(n)^2, x(0)=1/10.

Original entry on oeis.org

1, 37, 10693, 84699253, 12357810823725973, 36165967884042486031784215609813, 10542455454648216809579521558960739092388747823498758873614315093
Offset: 0

Views

Author

Reinhard Zumkeller, Dec 22 2012

Keywords

References

  • H. O. Peitgen and P. H. Richter, The Beauty of Fractals. Springer, 1986, p. 23f.

Crossrefs

Cf. A051840, A220812 (denominators).

Programs

  • Haskell
    import Data.Ratio ((%), numerator)
    a220811 n = a220811_list !! n
    a220811_list = map numerator vs where
       vs = iterate (\x -> x * (4 - 3 * x)) (1 % 10)
    -- Reinhard Zumkeller, Dec 22 2012

A225162 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 10/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

Original entry on oeis.org

1, 9, 91, 9181, 92480761, 9304615055139121, 93529710772930377727152664652641, 9394835719974970982728198049552322910011762062750179997188274881
Offset: 1

Views

Author

Martin Renner, Apr 30 2013

Keywords

Comments

Numerators of the sequence of fractions f(n) is A165428(n+1), hence sum(A165428(i+1)/a(i),i=1..n) = product(A165428(i+1)/a(i),i=1..n) = A165428(n+2)/A225169(n) = A220812(n-1)/A225169(n).

Examples

			f(n) = 10, 10/9, 100/91, 10000/9181, ...
10 + 10/9 = 10 * 10/9 = 100/9; 10 + 10/9 + 100/91 = 10 * 10/9 * 100/91 = 10000/819; ...
		

Crossrefs

Programs

  • Maple
    b:=n->10^(2^(n-2)); # n > 1
    b(1):=10;
    p:=proc(n) option remember; p(n-1)*a(n-1); end;
    p(1):=1;
    a:=proc(n) option remember; b(n)-p(n); end;
    a(1):=1;
    seq(a(i),i=1..8);

Formula

a(n) = 10^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.
a(n) = 10^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

A225169 Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 10/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

Original entry on oeis.org

1, 9, 819, 7519239, 695384944860879, 6470289227069622272847335347359, 605164280025029017271801950447677089988237937249820002811725119
Offset: 1

Views

Author

Martin Renner, Apr 30 2013

Keywords

Comments

Numerators of the sequence s(n) of the sum resp. product of fractions f(n) is A165428(n+2), hence sum(A165428(i+1)/A225162(i),i=1..n) = product(A165428(i+1)/A225162(i),i=1..n) = A165428(n+2)/a(n) = A220812(n-1)/a(n).

Examples

			f(n) = 10, 10/9, 100/91, 10000/9181, ...
10 + 10/9 = 10 * 10/9 = 100/9; 10 + 10/9 + 100/91 = 10 * 10/9 * 100/91 = 10000/819; ...
s(n) = 1/b(n) = 10, 100/9, 10000/819, ...
		

Crossrefs

Programs

  • Maple
    b:=proc(n) option remember; b(n-1)-b(n-1)^2; end:
    b(1):=1/10;
    a:=n->10^(2^(n-1))*b(n);
    seq(a(i),i=1..7);

Formula

a(n) = 10^(2^(n-1))*b(n) where b(n)=b(n-1)-b(n-1)^2 with b(1)=1/10.
Showing 1-4 of 4 results.