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 1-3 of 3 results.

A185332 Numerators of u(n) where u(n) = (u(n-1) + u(n-2)) / u(n-3), with u(1) = u(2) = u(3) = 1.

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 4, 3, 7, 11, 5, 29, 155, 224, 639, 3787, 1837, 2855, 32393, 97309, 127512, 1825907, 13672693, 12048382, 61067879, 1364331725, 942450221, 3863275086, 126646632559, 699156998051, 2194555785960, 61774183992421
Offset: 1

Views

Author

Michael Somos, Jan 27 2012

Keywords

Examples

			u(1), ... = 1, 1, 1, 2, 3, 5, 4, 3, 7/5, 11/10, 5/6, 29/21, 155/77, 224/55, 639/145, ...
		

Crossrefs

Programs

  • Mathematica
    Numerator[RecurrenceTable[{a[1]==a[2]==a[3]==1,a[n]==(a[n-1]+a[n-2])/ a[n-3]}, a,{n,40}]] (* Harvey P. Dale, Jan 28 2013 *)
  • PARI
    {a(n) = local(v = [1, 1, 1]); if( n<1, n = 4-n); if( n<4, 1, for( k=4, n, v = [v[2], v[3], (v[2] + v[3]) / v[1]]); numerator( v[3] ))};

Formula

a(4 - n) = a(n) for all n in Z.
0 = u(n) * u(n+3) - u(n+1) - u(n+2) for all n in Z. - Michael Somos, Nov 01 2014

A185341 Denominators of u(n) where u(n) = (u(n-1) + u(n-2)) / u(n-3), with u(1) = u(2) = u(3) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 5, 10, 6, 21, 77, 55, 145, 899, 868, 1988, 38411, 90347, 95357, 637807, 3506263, 2382501, 19519203, 649945741, 911672929, 3857971277, 130630182325, 366719420575, 764101349503, 12533062448579, 136235802233249
Offset: 1

Views

Author

Michael Somos, Jan 27 2012

Keywords

Examples

			u(1), ... = 1, 1, 1, 2, 3, 5, 4, 3, 7/5, 11/10, 5/6, 29/21, 155/77, 224/55, 639/145, ...
		

Crossrefs

Programs

  • Mathematica
    Denominator[RecurrenceTable[{u[1] == u[2] == u[3] == 1, u[n] == (u[n - 1] + u[n - 2])/u[n - 3]}, u, {n, 50}]] (* G. C. Greubel, Jun 27 2017 *)
  • PARI
    {u(n) = local(v = [1, 1, 1]); if( n<1, n = 4-n); if( n<4, 1, for( k=4, n, v = [v[2], v[3], (v[2] + v[3]) / v[1]]); denominator( v[3] ))};

Formula

u(4 - n) = u(n) for all n in Z.
0 = u(n) * u(n+3) - u(n+1) - u(n+2) for all n in Z. - Michael Somos, Nov 01 2014

A330080 a(n) = floor(b(n)), where b(1) = b(2) = b(3) = 1 and b(n) = (b(n-1) + b(n-2))/b(n-3) for n > 3.

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 0, 1, 2, 4, 4, 4, 2, 1, 0, 1, 1, 2, 3, 5, 3, 2, 1, 1, 0, 1, 2, 4, 4, 3, 1, 1, 0, 1, 1, 3, 4, 4, 2, 1, 0, 1, 1, 2, 3, 5, 3, 2, 1, 1, 0, 1, 2, 4, 4, 3, 1, 1, 0, 1, 1, 3, 4, 4, 2, 1, 0, 1, 1, 2, 3, 5, 3, 2, 1, 1, 0, 1, 2, 4, 4, 3, 1, 1, 0, 1, 1, 3, 4, 4, 2, 1, 0, 1, 1, 2, 3, 5
Offset: 1

Views

Author

Andres Cicuttin, Nov 30 2019

Keywords

Comments

This sequence seems quasiperiodic with the same quasiperiod of A185332(n)/A185341(n) (see related comments of Michael Somos in A068508).
Conjecture: 0 <= a(n) <= 5.

Crossrefs

Programs

  • Mathematica
    c[1] = 1; c[2] = 1; c[3] = 1;
    c[n_] := c[n] = (c[n - 2] + c[n - 1])/c[n - 3];
    Table[Floor@c[j], {j, 1, 2^6}]
Showing 1-3 of 3 results.