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.

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

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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, ...

Links

Crossrefs

Cf. A068508, A185341, A205303.

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

A205303 a(n) where a(n) * a(n-5) * a(n-10) = a(n-1) * a(n-6) * a(n-8) + a(n-2) * a(n-4) * a(n-9), with a(1) = ... = a(10) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 5, 8, 18, 21, 44, 60, 174, 372, 1344, 2556, 12984, 24048, 82224, 160848, 904032, 1967328, 14812992, 43671744, 374004864, 1108847232, 8442489600, 18677267712, 211090572288, 612702392832, 6883734979584
Offset: 1

Views

Author

Michael Somos, Jan 28 2012

Keywords

Comments

The recursion has the Laurent property. If a(1), ..., a(10) are variables, then a(n) is a Laurent polynomial -- a rational function with a monomial denominator.
Similar to the Somos-5 sequence, the sequence a(n) can be expressed in terms of the Jacobi theta_3(u, q) function as a(n) = c1 * c2^(n - c6)^2 * theta_3(c4*n - c5, c3) where both c1 and c5 depend on the residue class of n modulo 12, c5 linearly with slope 0.2347354... with c5 = 0.4030547... if n=12*k+6, c6 = 5.5 + (-1)^n * 0.1844232..., c2 = 1.0303784..., c4 = 0.6231417..., q = c3 = 0.116755251... = exp(Pi i tau) and 3 * (72961 / 432)^3 / 1367 = 10572.4060... the corresponding invariant j(tau).

Links

Crossrefs

Cf. A185332, A185341.

Programs

  • Mathematica
    nxt[{a10_,a9_,a8_,a7_,a6_,a5_,a4_,a3_,a2_,a1_}]:={a9,a8,a7,a6,a5,a4,a3,a2,a1,(a1*a6*a8+a2*a4*a9)/(a5*a10)}; Transpose[ NestList[ nxt,Table[1, {10}],40]][[1]] (* Harvey P. Dale, Mar 27 2015 *)
a[ n_] := Which[ n < 6, a[11 - n], n < 11, 1, True, (a[n - 1] a[n - 6] a[n - 8] + a[n - 2] a[n - 4] a[n - 9]) / (a[n - 5] a[n - 10])]; (* Michael Somos, Oct 21 2018 *)
a[ n_] := Which[ n < 6, a[11 - n], n < 11, 1, n < 13, n - 9, True, (-a[n - 1] a[n - 12] + 13 a[n - 4] a[n - 9]) / a[n - 13]]; (* Michael Somos, Oct 21 2018 *)
  • PARI
    {a(n) = my(v); if( n<1, n = 11-n); v = vector( n, k, 1); for( k=11, n, v[k] = (v[k-1] * v[k-6] * v[k-8] + v[k-2] * v[k-4] * v[k-9]) / (v[k-5] * v[k-10])); v[n]};
  • PARI
    {a(n) = my(v); if( n<1, n = 11-n); v = vector( n, k, 1); for( k=11, n, v[k] = ( -v[k-3] * v[k-4] + v[k-1] * v[k-6] * [2, 2, 2, 3] [k%4 + 1]) / v[k-7]); v[n]};
  • PARI
    {a(n) = my(v); if( n<1, n = 11-n); v = vector( n, k, 1); for( k=11, n, v[k] = ( v[k-1] * v[k-4] * [3, 3, 4] [k%3 + 1] - v[k-2] * v[k-3] * [3, 2, 2, 2] [k%4 + 1]) / v[k-5]); v[n]};

Formula

Let u(n) := (a(n) * a(n+7)) / (a(n+3) * a(n+4)) = A185332(n) / A185341(n), then u(n) = (u(n-1) + u(n-2)) / u(n-3), u(1) = u(2) = u(3) = 1.
a(n) = a(11-n) for all n in Z.
a(n+7) * a(n-6) = -a(n+6) * a(n-5) + 13 * a(n+3) * a(n-2) for all n in Z. [see Grammaticos et al., Equation (3.2) for the general form of this equation.]
a(n+7) * a(n-7) = a(n+5) * a(n-5) + 13 * a(n+1) * a(n-1) for all n in Z.
a(n+11) * a(n-11) = 156*a(n+5) * a(n-5) + 612 * a(n+1) * a(n-1) for all n in Z. - Michael Somos, Oct 19 2023
a(n+3) * a(n-2) = (3 + [3|(n+1)]) * a(n+2) * a(n-1) - (2 + [4|n]) * a(n+1) * a(n) for all n in Z where [] is the Iverson bracket. - Michael Somos, Oct 19 2023

A068508 a(n) = round((a(n-1) + a(n-2))/a(n-3)) starting with a(1)=a(2)=a(3)=1.

Original entry on oeis.org

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

Views

Author

Henry Bottomley, Mar 25 2002

Keywords

Comments

While this sequence has period 8, the unrounded version b(n) = (b(n-1) + b(n-2))/b(n-3) seems to have a quasi-period of about 8.7 for this particular starting point.
The unrounded version b(n) = A185332(n) / A185341(n) as given in A205303 has 8.694171... quasi-period. - Michael Somos, Oct 22 2018

Examples

			a(7) = round((a(6) + a(5))/a(4)) = round((5+3)/2) = 4.

Links

Crossrefs

Cf. A048112, A185332, A185341, A205303.

Formula

a(n) = a(n-8).

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

Cf. A068508, A185332, A185341.

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 1-4 of 4 results.

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